TITLE: Degree of a map $T^2\to T^2$ induced by an $2\times 2$ integer matrix
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.