The second homology group of the universal cover of $S^2\vee S^1\vee S^1$ is nontrivial Let $E$ be a universal cover of $S^1\vee S^1\vee S^2$ which exists. I want to show $H_2(E)\neq 0$. One of the solution given in this pdf (last page) is that the inclusion $i:S^2\hookrightarrow S^2\vee S^1\vee S^1$ of the first wedge summand factors through the universal covering space. Why this is true? Could you explain this?
 A: The universal cover is a Serre fibration, so has homotopy lifting property. Now stare at https://en.wikipedia.org/wiki/Fibration#/media/File:Fibration_homotopy_groups_LES_connecting_morphism_diagram.svg
for $n=2$.
Another way to say this, is that we can view the inclusion of $S^2$ into $S^2\vee S^1 \vee S^1$ as a homotopy of the constant loop at $i(\text{north pole})$ to the constant loop at $i(\text{south pole})$ . We lift the boundary to the basepoint, and, by homotopy lifting property, get a lift - a map of the cylinder $S^1\times[0,1]$ to the universal cover. By construction, $S^1\times{0}$ is mapped to the basepoint; the $S^1\times{1}$ is mapped to the fiber over $i(\text{south pole})$; but that fiber is discrete, so $S^1\times{1}$ is also mapped to a single point. Thus the lifted map of $S^1\times[0,1]$ descends to a map of $S^2$.  This is the lift of $i$.
A: The inclusion factors through the universal cover by covering space theory as follows. For any covering space $p:(\tilde{X},\tilde{x_0})\to (X,x_0)$ and any map $f:(Y,y_0)\to (X,x_0)$, we get that a lift of $f$ along $p$ exists if and only if $f_*(\pi_1(Y,y_0))\subset p_*(\pi_1(\tilde{X},\tilde{x_0})) $. This is Proposition 1.33 in Hatcher. Since $S^2$ is simply-connected, we get the wanted lift.
To show that $H_2(E)\neq 0$, we apply additivity of reduced homology. This says that the wegde of inclusions induces a homology isomorphism $H_2(S^2)\oplus H_2(S^1)\oplus H_2(S^1)\cong H_2(S^2\vee S^1 \vee S^1)$. We then get that the inclusion induces isomorphism $H_2(i):H_2(S^2)\cong H_2(S^2\vee S^1 \vee S^1)$. This factors through $H_2(E)$ by the lifting hence $H_2(E)$ must be nontrivial.
