How to determine the map between the singular homology groups of a torus induced by a linear map $f:\mathbb R^2 \to \mathbb R^2$? Let $f:\mathbb R^2 \to \mathbb R^2$ be the linear map whose matrix under the standard basis is $ \begin{bmatrix}
2 &amp; 1 \\
1 &amp; 1 
\end{bmatrix} $.

Clearly $f$ induces the map from $\mathbb R^2/ \mathbb Z^2$ to itself. I wonder how to find the induced map between homology groups of the torus. 

There are three nontrivial homologies: $H_0(T^2)=\mathbb Z, H_1(T^2)=\mathbb Z^2, H_2(T^1)=\mathbb Z$. Since $f: \mathbb R^2/\mathbb Z^2 \to \mathbb R^2/\mathbb Z^2$ is an isomorphism, by the functorality, so are $f_*: H_*(T^2)\to H_*(T^2)$. 

But I can't figure out if $H_i(T^2)\to H_i(T^2)$ are the identity map or $-1$ maps when $i=0,2$. For $i=1$, I have a feeling that the induced homology map $f_*$ sends $(x,y)\in \mathbb Z^2$ to $(2x+y,x+y)$ (natively induced by the matrix), but I don't know how to show it rigorously.

I hope the method is not too advanced (best if it only uses the definition).
The best way to understand the map in $H_0$ is to consider this map as a map of CW-complexes. Than marked point is mapped to itself, so the map is an isomorphism.

If a map in $H_2$ would be multiplication by $-1$ it would mean that the map does not preserve the orientation. But your map does preserve it since it preserves the orientation of a plane(because the determinant is equal to 1). So the map on $H_2$ is also an identity.
The best way to understand the map in $H_0$ is to consider this map as a map of CW-complexes. Than marked point is mapped to itself, so the map is an isomorphism.

If a map in $H_2$ would be multiplication by $-1$ it would mean that the map does not preserve the orientation. But your map does preserve it since it preserves the orientation of a plane(because the determinant is equal to 1). So the map on $H_2$ is also an identity.