Induced map on homology from a covering space isomorphism Suppose $S^1 \times \mathbb{R}P^2$ covers some space.  Why is it that any covering space isomorphism $h$ induces the identity map on $H_1$?  I don't see how to prove this except maybe from looking at the explicit construction of the map from $\pi_1$ to $H_1$. 
 A: A covering transformation is in particular an element of ${\rm Homeo}(S^1\times\mathbb{RP}^2)$, so it induces an automorphism of homology $\varphi\in {\rm Aut}(\mathbb Z\times \mathbb Z_2)$. Now it is an easy exercise to show that $ {\rm Aut}(\mathbb Z\times \mathbb Z_2)$ consists of the identity map, $(a,b)\mapsto (-a,b)$, $(a,b)\mapsto (-a,b+a)$ and $(a,b)\mapsto (a,b+a)$. So we just have to rule these last three maps out. Suppose $f$ is a covering transformation inducing one of these maps. Now we use the fact that covering transformations cannot have fixed points and the Lefschetz fixed point theorem comes to the rescue. In the two maps where the coefficient of $a$ is negative, the alternating sum of traces of $f$ on the rational homologies $H_0\cong \mathbb Q$ and $H_1\cong\mathbb Q$ is $1-(-1)\neq 0$. So $f$ has a fixed point, which contradicts the possibility that it is a covering transformation. Hence we only need to rule out $(a,b)\mapsto (a,b+a)$. Unfortunately I don't see how to do that at the moment...
I'd like to add that this question depends on the particular space $S^1\times\mathbb{RP}^2$ and not just its fundamental group, as the following example shows. Let 
$$G=\langle a,b,c|[a,b]=1, b^2=1, cac^{-1}=ab, [c,b]=1\rangle.$$
The subgroup generated by $a,b$ is a normal subgroup. It is clear that it is a quotient of $\mathbb Z\oplus\mathbb Z_2$, but in fact you can show that it is isomorphic to this group (see below). So let $X$ be a space with $\pi_1(X)=G$, and consider the normal cover $Y$ with fundamental group $\mathbb Z\oplus\mathbb Z_2$ corresponding to this subgroup. Then the deck transformation group is isomorphic to $\mathbb Z$ generated by $c$ and it acts on $H_1(Y)$ by sending $a$ to $a+b$ and $b\mapsto b$.
To see that the subgroup generated by $a$ and $b$ is isomorphic to $\mathbb Z\oplus\mathbb Z_2$, we give a different description of $G$. Notice that $b$ commutes with everything and $ca=acb$, so in a word representing an element in $G$, we can always rearrange it to be of the form $a^kc^\ell b^m$, with $a^kc^\ell b^m\cdot a^uc^v b^w=a^{k+u}c^{\ell+v}b^{m+w+\ell u}$. So $G$ is isomorphic to $\mathbb Z\times\mathbb Z\times \mathbb Z_2$ with a funny group multiplication: $(x,y,z)*(x',y',z')=(x+x',y+y',z+z'+y\cdot x').$ One can see that $\mathbb Z\times\{0\}\times \mathbb Z_2$ forms a subgroup isomorphic to $\mathbb Z\times\mathbb Z_2$.
