Let $f$ be a map from $X= S^1 \times S^1$ to itself that is the identity on one factor and a reflection on the other. Then how does the induced map $f_* :H_2(X)\to H_2(X)$ look like?
Since $f$ is just a reflection, I guess $f_* = -\mathbb{1}$, but then I have no idea how I can prove it. I encountered this problem when I was solving exercises in Hatcher, where I had to compute the homology groups of the mapping torus $T_f$ of $f$. In Hatcher, we have the long exact sequence $$ 0\to H_3(T_f) \to H_2(X) \xrightarrow{\mathbb{1} -f_*} H_2(X) \to H_2(T_f) \to H_{1}(X) \to \ldots $$ so I have to know the map $\mathbb{1}-f_*$.
For spheres, we have a very simple simplicial structure on it, so it is easy to calculuate the degree of the reflection map, but how do I calculate it for general complexes like $S^1 \times S^1$? I tried to give a symmetric-looking simplicial structure on $S^1 \times S^1$, but it requires too many simplices.
I also have no idea for the induced map on $H_1(X)$, but I guess I can get some idea if I know the case of second homology group.