Degree of a map $T^2\to T^2$ induced by an $2\times 2$ integer matrix

Question

Note that the $2$-torus $T^2$ can be seen as a quotient space $\Bbb R^2/\Bbb Z^2$ of $\Bbb R^2$. Then any $2\times 2$ integer matrix $A=(\begin{smallmatrix} a & b\\ c & d \end{smallmatrix})$ gives a well-defined map $A:T^2\to T^2$. On the other hand, we have $H_1(T^2)=\Bbb Z^2$ and $H_2(T^2)=\Bbb Z$. What I want to show is, the map $A_*:H_2(T^2)\to H_2(T^2)$ induced by $A$, is given by $\Bbb Z\xrightarrow{\times \det(A)} \Bbb Z$, multiplication by $\det(A)$, and the map $A_*:H_1(T^2)\to H_1(T^2)$ is given by $\Bbb Z^2 \xrightarrow{A} \Bbb Z$.

Actually I want to use this result in Exercise 30 of section 2.2, in Hatcher's Algebraic Topology. (http://pi.math.cornell.edu/~hatcher/AT/AT.pdf) Parts (c) to (e) would become easy then.

The map on $H_1$ seems to be computed if we use the identification $\pi_1(T^2)=H_1(T^2)$ (which is after section 2.2, though), but I have no idea for the map on $H_2$. (Maybe a local degree argument?, but I'm not sure) Thanks in advance.

Answer 1

There's a natural map $\wedge^2(H^1(T^2, \mathbb{Z})) \to H^2(T^2, \mathbb{Z})$ given by the cup product. Show that it's an isomorphism. Then the desired result follows from the fact that $\det(A)$ is the scalar by which $A$ acts on the top exterior power, together with the universal coefficient theorem.

Answer 2

If you use deRham cohomology, it's pretty clear that the area-form (which is a generator for $H^2$) gets multiplied by the determinant --- that's what the change of variables formula from calculus says.

Now use the fact that $H_2(T, \Bbb R) = hom( H^2(T, \Bbb R) ) $ to conclude that the corresponding map on $H_2$ must also be multiplication by $det$.

As an alternative, consider the 0-cocycle $\mu$ defined by "counting (signed) points". Because $T$ is connected, $\mu$ is clearly a generator. A drawing of the map $A$ shows that it takes any irrational point, like $(\pi/7, e/3)$ to $k = det(A)$ separate points, so the induced map on $H^0$ is multiplication by $k$. Poincare duality then makes the induced map on $H_n$ be multiplication by $k$ as well.