I want to show, that although the homology groups of $X :=S^1 \vee S^1 \vee S^2$ and the torus $T^2$ are isomorphic, the homology groups of their universal covers are not. Let $U_X$ be the universal cover of $X$
The first part was easy since $\tilde{H}_k(Y_1 \vee Y_2) = \tilde{H}_k(Y_1) \oplus \tilde{H}_k(Y_2)$. The universal cover of $T^2$ is $\Bbb R^2$, since $\Bbb R$ is the universal cover of $S^1$. Therefore $H_k(\Bbb R^2)=0$ if $k>0$ and $H_0(\Bbb R^2)=\Bbb Z$.
Of course computing $H_k(U_X)$ is a way to solve this problem, but I don't know how compute $U_X$. Also my tutor told me, that this is not necessary and the properties of the universal cover are sufficient to solve this.
my thoughts so far
I think, that showing $H_2(U_X)\not= 0$ is now the best way to proof the statement, because $H_2(S^2)= \Bbb Z$ and $S^2$ is the universal cover of $S^2$ and has to be "contained" in $U_X$ somehow…
Let $p \colon U_X \to X$ the covering map. I tried to show that the induced map $p_* \colon H_2(U_X) \to H_2(X) \cong \Bbb Z$ has non-trivial image (or even is surjective) by looking at the pair sequences of $(U_X,A)$ and $(X,B)$, where $A \subset U_X$ is one of the disjoint subsets of $p^{-1}(B)$, but that did not work out.
Thanks in advance for any help 🙂
PS: I already read this, but it didn't help me for the reasons mentioned above.
Best Answer
The universal cover of $T$ is contractible hence homology vanishes. Now consider $\hat X$ the universal cover. You know that $H_2(X) \neq 0$ and you also know that you have the inclusion $i: S^2 \to X$ and that $H_2(i):H_2(S^2)\to H_2(X)$ is non-trivial. But $S^2$ is simply connected, hence there is a lift $\tilde i:S^2 \to \hat X$ and since $H_2(i)=H_2(p\tilde i)=H_2(p)H_2(\tilde i)$ is non trivial as mentioned before, so is $H_2(\tilde i)$. In particular $\hat X$ has non-trivial homology.