This question is related to my old question: Universal cover of $(S^1\times S^2)\vee S^1$. How can we compute the second homology group $H_2(E)$, where $E$ is the universal cover of $(S^1\times S^2)\vee S^1$? I attempted as follows: Since the base space $B=(S^1\times S^2)\vee S^1$ is a CW complex, $E$ also admits a CW complex structure, and its $1$-skeleton $E^1$ is the universal cover of $S^1\vee S^1$ (a picture is in the link), which is a simply connected graph, and thus contractible. So if we consider the long exact sequence of the pair $(E,E^1)$, it follows that $H_2(E)$ is isomorphic to $H_2(E,E^1)$, which is isomorphic to $H_2(E/E^1)$. But I think $H_2(E/E^1)$ cannot be easily computed, and I got stuck.
Second homology group of the universal cover of $(S^1\times S^2)\vee S^1$
algebraic-topologycovering-spacescw-complexeshomology-cohomology
Related Solutions
This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.
Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~ \underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~ \underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram $$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\ &\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\ 0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0 \end{array}$$ Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.
Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\ &\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\ 0\to& \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0 \end{array}$$
From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.
Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map
$\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.
As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$
To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence $$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$
Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$
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.
Best Answer
The 1-skeleton is a tree, so is contractible. This leaves us with something homotopy equivalent to a infinite wedge of spheres (if I am understanding the cover correctly), the second homology of such a thing is an infinite direct sum of the integers.