I am having trouble understanding how to compute the induced map for the second homology. For example say I have $\varphi:\mathbb{T}^2\rightarrow \mathbb{T}^2$ that is a self homeomorphism, then what would be the general strategy to compute $\varphi_*:H_2(\mathbb{T}^2)\rightarrow H_2(\mathbb{T}^2)$.
I would appreciate if someone could work through a non trivial example. Thank you in advance.
Best Answer
There are several different ways of computing the induced map on $H_2$. For example, one can give $\mathbb{T}^2$ a CW- complex structure. It's a general theorem that every map of CW complexes is homotopic to a CW-map (one which maps the $k$-skeleton to the $k$-skeleton), and that homotopic maps induce the same map on homology. One your map is CW, it's easy (or at least, easier) to compute induced maps.
One can also use a simplicial decomposition, etc.
Another approach is to factor your map $f:X\rightarrow Y$ as a composition of maps $X\rightarrow Y \rightarrow Z$ whose induced maps you know.
A third approach is to try to understand induced maps on cohomology and then relate that back to homology. The main advantange here is that the cohomology groups fit together to have a natural ring structure, and $\phi^\ast$ ends up being a ring homomorphism. Thus knowledge of $\phi^\ast$ on $H^1$ often (but not always) informs one what $\phi^\ast$ is doing on $H^2$.
Finally, in the case where the dimension of your space is $2$ and your space is nice enough (like an orientable manifold), then degree theory helps to answer the question.
In your case, since your specified that $\phi$ is a homeomorphism, it follows that $\phi_\ast$ is an isomorphism, and the only isomorphisms of $\mathbb{Z} = H_2(\mathbb{T}^2)$ are multiplication by $\pm 1$. It follows that $\phi_\ast(x) = x$ or $\phi_\ast(x) = -x$. (Without more information, we can't tell which - there are maps realizing either $\phi_\ast$).