Necessary condition on homology group for a set to be contractible The Next CEO of Stack OverflowAlgebraic TopologyWhat is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?Fundamental group of the Poincaré Homology SphereNecessary condition for removing a simplex and changing homotopy type.Does trivial fundamental group imply contractible?Homology of Eilenberg-MacLane $K(pi,1)$ in terms of group homology and TorHomology of contractible spaceFundamental group generators of null homology manifoldsHomology group of $mathbbS^1 vee mathbbRP^2$ and covering spacesDo contractible homology manifolds have one end?

Is French Guiana a (hard) EU border?

How is this set of matrices closed under multiplication?

Where do students learn to solve polynomial equations these days?

Writing differences on a blackboard

Can you be charged for obstruction for refusing to answer questions?

Why did CATV standarize in 75 ohms and everyone else in 50?

Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?

Easy to read palindrome checker

Why does standard notation not preserve intervals (visually)

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

What was the first Unix version to run on a microcomputer?

What did we know about the Kessel run before the prequels?

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

How did people program for Consoles with multiple CPUs?

Is it professional to write unrelated content in an almost-empty email?

Running a General Election and the European Elections together

Are police here, aren't itthey?

How to edit “Name” property in GCI output?

How do I align (1) and (2)?

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

Why isn't acceleration always zero whenever velocity is zero, such as the moment a ball bounces off a wall?

Won the lottery - how do I keep the money?

The past simple of "gaslight" – "gaslighted" or "gaslit"?

How to prove a simple equation?



Necessary condition on homology group for a set to be contractible



The Next CEO of Stack OverflowAlgebraic TopologyWhat is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?Fundamental group of the Poincaré Homology SphereNecessary condition for removing a simplex and changing homotopy type.Does trivial fundamental group imply contractible?Homology of Eilenberg-MacLane $K(pi,1)$ in terms of group homology and TorHomology of contractible spaceFundamental group generators of null homology manifoldsHomology group of $mathbbS^1 vee mathbbRP^2$ and covering spacesDo contractible homology manifolds have one end?










3












$begingroup$


We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



    Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



    I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



    Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
    Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



      Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



      I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



      Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
      Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










      share|cite|improve this question









      $endgroup$




      We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



      Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



      I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



      Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
      Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?







      algebraic-topology simplicial-complex






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      Sanae KochiyaSanae Kochiya

      626




      626




















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



          It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



          By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



          However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              Thank you for your response. Do you mind direct me to the proof of your statement?
              $endgroup$
              – Sanae Kochiya
              57 mins ago










            • $begingroup$
              For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
              $endgroup$
              – hunter
              12 mins ago


















            0












            $begingroup$

            This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



            But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168982%2fnecessary-condition-on-homology-group-for-a-set-to-be-contractible%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



              It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



              By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



              However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                  It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                  By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                  However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






                  share|cite|improve this answer









                  $endgroup$



                  The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                  It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                  By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                  However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 39 mins ago









                  WilliamWilliam

                  2,9351224




                  2,9351224





















                      2












                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        57 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        12 mins ago















                      2












                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        57 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        12 mins ago













                      2












                      2








                      2





                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$



                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      hunterhunter

                      15.4k32640




                      15.4k32640











                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        57 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        12 mins ago
















                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        57 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        12 mins ago















                      $begingroup$
                      Thank you for your response. Do you mind direct me to the proof of your statement?
                      $endgroup$
                      – Sanae Kochiya
                      57 mins ago




                      $begingroup$
                      Thank you for your response. Do you mind direct me to the proof of your statement?
                      $endgroup$
                      – Sanae Kochiya
                      57 mins ago












                      $begingroup$
                      For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                      $endgroup$
                      – hunter
                      12 mins ago




                      $begingroup$
                      For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                      $endgroup$
                      – hunter
                      12 mins ago











                      0












                      $begingroup$

                      This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                      But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                        But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                          But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                          share|cite|improve this answer









                          $endgroup$



                          This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                          But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 33 mins ago









                          Connor MalinConnor Malin

                          584111




                          584111



























                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168982%2fnecessary-condition-on-homology-group-for-a-set-to-be-contractible%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              How should I use the fbox command correctly to avoid producing a Bad Box message?How to put a long piece of text in a box?How to specify height and width of fboxIs there an arrayrulecolor-like command to change the rule color of fbox?What is the command to highlight bad boxes in pdf?Why does fbox sometimes place the box *over* the graphic image?how to put the text in the boxHow to create command for a box where text inside the box can automatically adjust?how can I make an fbox like command with certain color, shape and width of border?how to use fbox in align modeFbox increase the spacing between the box and it content (inner margin)how to change the box height of an equationWhat is the use of the hbox in a newcommand command?

                              Doxepinum Nexus interni Notae | Tabula navigationis3158DB01142WHOa682390"Structural Analysis of the Histamine H1 Receptor""Transdermal and Topical Drug Administration in the Treatment of Pain""Antidepressants as antipruritic agents: A review"

                              inputenc: Unicode character … not set up for use with LaTeX The Next CEO of Stack OverflowEntering Unicode characters in LaTeXHow to solve the `Package inputenc Error: Unicode char not set up for use with LaTeX` problem?solve “Unicode char is not set up for use with LaTeX” without special handling of every new interesting UTF-8 characterPackage inputenc Error: Unicode character ² (U+B2)(inputenc) not set up for use with LaTeX. acroI2C[I²C]package inputenc error unicode char (u + 190) not set up for use with latexPackage inputenc Error: Unicode char u8:′ not set up for use with LaTeX. 3′inputenc Error: Unicode char u8: not set up for use with LaTeX with G-BriefPackage Inputenc Error: Unicode char u8: not set up for use with LaTeXPackage inputenc Error: Unicode char ́ (U+301)(inputenc) not set up for use with LaTeX. includePackage inputenc Error: Unicode char ̂ (U+302)(inputenc) not set up for use with LaTeX. … $widehatleft (OA,AA' right )$Package inputenc Error: Unicode char â„¡ (U+2121)(inputenc) not set up for use with LaTeX. printbibliography[heading=bibintoc]Package inputenc Error: Unicode char − (U+2212)(inputenc) not set up for use with LaTeXPackage inputenc Error: Unicode character α (U+3B1) not set up for use with LaTeXPackage inputenc Error: Unicode characterError: ! Package inputenc Error: Unicode char ⊘ (U+2298)(inputenc) not set up for use with LaTeX