These days I'm really interested in drawing pictures.Please look at the picture below,I think it is a contractible topology space(The boundary is $abcd$,and the interior of $abcd$ is $\textbf{included}$!).It's well known that Euclidean space $\mathbb{R}^n$ and convex set in $\mathbb{R}^n$ is contractible.Then how about the set $S \subseteq \mathbb{R}^n$ which is not convex?(Hope to get some typical examples...) -- Question $1$.

And question $2$ is about the geometric intuition.Let $X$ be a topological space.Is homology $H_0(X)=\mathbb{Z}$ and $H_i(X) = 0,i \in N_+$ implies $X$ contractible?--Question $2$.

These questions come from myself.As for question $1$ some non-convex set is contractible such as the picture shown and some are not just like $S^n$ is $R^{n+1}$.As for question $2$ just my intuition,however I find no tools to prove it.I always get useful advice on this website.

Math is my sub-degree...I really need a intuitive explanation and some examples,thank you very much.

enter image description here

  • $\begingroup$ Here is an answer to question 2 : math.stackexchange.com/questions/29957/… $\endgroup$ – user171326 Jun 16 '17 at 13:39
  • $\begingroup$ Nice to see you again!Could you give a sketch on it? $\endgroup$ – Muse_China Jun 16 '17 at 13:40
  • $\begingroup$ Sure, this is the answer given by the user "MartianInvader". One can construct a space which has non-trivial fundamental group but trivial homology, by constructing a group with $\pi_1, \pi_2$ perfect group (this means that the abelianization of the groups is zero) so in particular $H_k = 0 \forall k > 0$. There are some details to fill I'm not sure I could do all of it. But in fact, if you can read french there is a very nice website which tried to describe how Poincaré was thinking of topology, this may be useful (sadly only in french) : analysis-situs.math.cnrs.fr $\endgroup$ – user171326 Jun 16 '17 at 13:50
  • $\begingroup$ So If a foundamental group $\pi_1$ is given,how can we construct the topology structure?(Maybe it's similar to Galois inverse problem which is a really big problem)Consider the foundamental group of $nT^2$ is $<a_1,b_1,a_2,b_2,...,a_n,b_n \mid a_1b_1a_1^{-1}b_1^{-1}...a_nb_na_n^{-1}b_n^{-1}>$.The subgroup $G$ of free group $<a_1,b_1,a_2,b_2,...,a_n,b_n>$ generated by $a_1b_1a_1^{-1}b_1^{-1}...a_nb_na_n^{-1}b_n^{-1}$ is perfect,however how can we construct a topological space whose fundamental group isomorphic to $G$? $\endgroup$ – Muse_China Jun 16 '17 at 14:03
  • $\begingroup$ Yes, for any group $G$ there is a space $X$ with $\pi_k(X) = 0$ if $k>1$ and $\pi_1(X) \cong G$. This is called a $K(G,1)$ space.(But if $H_1(X) = 0$ I'm not sure it implies that $H_k(X) = 0$ for $k > 1$ ...) $\endgroup$ – user171326 Jun 16 '17 at 14:15

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