Product of maps and homology groups Let $f_1:S^1 \to S^1$ be a map of degree $2$ (i.e., $f_1(z)=z^2$ when we choose a proper coordinate centered at the origin) and $f_2:S^1 \to S^1$ be a map of degree $3$ ($f_2(z)=z^3$). Now define a map $g:S^1\times S^1 \to S^1\times S^1$ by putting $g(z,w):=(f_1(z),f_2(w))$. Then, how can I describe explicitly the induced maps between these homology groups $g_i':H_i(T^2)\to H_i(T^2)$ for $i=0,1,2$? I guess $g_1'(n,m)=(2n,3m)$ and the others are the identities, but I cannot prove it rigorously.
 A: OK. If you know cellular homology, you can look at the representation of the torus as the unit square in the plane (with opposite edges of the square pairwise identified), thinking of the rest of the plane as being divided up into a grid of unit squares by the integer lattice. 
Now the map you've got sends the x-axis to itself by $x \mapsto 2x$ and the $y$ axis to itself by $ \mapsto 2y$. The unit square is a cell that generates the 2nd homology group, right? Well, under this map, it gets sent to 6 copies of itself. What's the induced map on homology? 
Alternatively, think of two cellular subdivisions of the unit square: one into a $3 \times 2$ array of small squares; the generator for 2nd homology is the sum of these six small squares. In the other one, there's just one big square. Now your map takes the generator of homology in the first cellulation into 6 times the generator of homology in the second. So you can explicitly write down the chain map; what's the induced map on homology? 
