Degree of a map from $ T^2 \rightarrow T^2$

Question

Given a map $f:T^2 \rightarrow T^2$, the degree of $f$ is given by the induced homomorphism $f^*:H_2(T^2) \rightarrow H_2(T^2)$.

If I know the induced homomorphism $f^{**}:H_1(T^2) \rightarrow H_1(T^2)$ is $f(x+y) = mx + ny$ (since $H_1(T^2) = \mathbb{Z} + \mathbb{Z}$) can I calculate the degree of the map?

Answer 1

Just to expand the comments into a complete answer:

1. $T^2$ is a $K(\mathbb Z^2,1)$, so basepoint-preserving homotopy classes of maps $T^2\to T^2$ are in bijective correspondence with fundamental group homomorphisms $\mathbb Z^2\to \mathbb Z^2$.

2. Since $\mathbb Z^2$ is abelian, the homotopy classes of maps are in bijective correspondence with homomorphisms $H_1(T^2)\to H_1(T^2)$.

3. If $f_*:H_1(T^2)\to H_1(T^2)$ is defined by $f_*(v)=Av$, with respect to some basis of $H_1(T^2)$, and with $A$ a $2\times 2$ integer matrix, then we may as well let $f$ be $f([x])=[Ax]$, where $x\in\mathbb R^2$ is a point in the universal cover, and where $[x]$ denotes the corresponding image in $T^2$. This $f$ induces the same map on $H^1$, so it must be homotopic to whatever the original $f$ was.

4. If $A$ is singular, then the degree is $0$. Otherwise, $\lvert \det A\rvert $ gives $\lvert f^{-1}(x_0)\rvert$, and the sign of $\det A$ gives the local degree ($\pm 1$), so $\det A$ is the degree.