Deck transformations on $S^1\times \mathbb{R}P^2$ I'm studying for qualifying exams and stuck on the following problem:
Suppose that $S^1\times \mathbb{R}P^2$ covers a space, and let $h$ be a deck transformation of the covering. Show that the induced isomorphism $h_*$ of $H_1(S^1\times \mathbb{R}P^2)$ is the identity.
I have made several attempts to solve it with no success. For example, if we can show that $h$ has a fixed point, we will be done. Alternatively, by looking at $\operatorname{Aut}(H_1(S^1\times \mathbb{R}P^2))\cong \mathbb{Z}_2^2$ and applying Lefschetz, we can show that $h_*$ is neither automorphism that maps $(1,0)\mapsto (-1,0)$. However, it remains to eliminate the possibility that $h_*$ is the automorphism that maps $(1,0)\mapsto (1,1)$ and $(0,1)\mapsto (0,1)$. Will either of these methods work? Does anyone know of a slicker proof?
 A: Suppose that the automorphism you are worried about is induced by a homeomorphism $h\in Homeo(S^1\times RP^2)$. Then a lift of $h$ to the universal cover conjugates an orientation-preserving deck-transformation to an orientation-reversing one, which is impossible. Hence, you are done.
Edit: Set $M=S^1\times RP^2$. I will identify $\pi_1(M)$ with the group of automorphisms of the universal cover $\tilde M\to M$. The group $\pi_1(M)$ splits as the product $Z\times Z_2$; let $a, b$ denote respective generators of the two direct factors. Then $a$ preserves orientation (of the universal cover $\tilde M$) while $b$ reverses orientation. In particular, $a+b$ also reverses the orientation. Therefore, a homeomorphism of $\tilde M$ cannot conjugate $a$ to $a+b$ (conjugation is understood in the group $Homeo(\tilde M)$). This is a general fact about homeomorphisms of oriented manifolds $X$: For $f_i\in Homeo(X)$, $i=1,2$, the homeomorphism 
$$
f_1 f_2 f_1^{-1} 
$$
preserves orientation of $X$ if and only if $f_2$ does. 
It follows that every $h\in Homeo(M)$ has to preserve the above direct product decomposition of $\pi_1(M)\cong H_1(M)$. 
