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 $\mathbb{Q}$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $\mathbb{P}^2$. When using $\mathbb{Z}$ 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?
Best Answer
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 $H_n(X;\mathbb{Z})=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.