Constructing a map of degree 2 $f:T^2\rightarrow S^2$ I know the definition of degree and homology type stuff. But I don't know what a map $T^2\rightarrow S^2$ should actually look like. We never work with explicit examples in my class and I just have no idea what to write. Should I map toroidal coordinates to toroidal coordinates? If not, then what? How do I build this map explicitly? I know his is a stupid question but please help, I'm having trouble with the formalism and I need somebody to directly spell it out for me. Please show me what to do. I don't know how to easily describe maps between spaces like this or how to tell intuitively what I should do to make it have a particular degree. I get what this doing algebraically to the homology groups but I don't know what it is doing to the actual set. 
 A: Consider the construction of the torus T as a square Q with some identifications on the boundary B. If you look at the image in T of the boundary of Q, you find two circles joined at a point. Call A the resulting subspace of T.
Now if you collapse A in T to a point to get the space T/A, you get the same thing as if you had collapsed the boundary B of the square Q to a point. That is, T/A is homeomorphic to Q/B, and this last space is a 2-sphere.
We get in this way a quotient map $f:T \to T/A\cong S^2$. This map has degree 1, as you can probably check.
Now if we identify $S^2$ with the extended complex plane, the map $q:z\mapsto z^2$ has degree 2.
The composition $q\circ f$ has degree 2.
A: Take any degree 1 map $S^1\times S^1 \to S^2$, (e.g. the map $S^1 \times S^1 \to S^1 \wedge S^1=S^2$) and compose it with the map $S^2 \to RP^2 \to RP^2/RP^1=S^2$ which has degree 2.  This is a degree 2 map.
A: You want a map $f: S^1  \times S^1 \to S^2$ of degree two. It suffices to provide a map of degree one, since the map $z \times z^2$ is of degree two from the torus to the torus, and degree is functorial.
For a map of degree one, consider the smash product map
$$\gamma: S ^1 \times S^1 \to S^1 \wedge S^1 \cong S^2.$$
To see that this has degree one, you can use the fact that cellular maps are determined by their degrees when collapsing subcomplexes. With the "standard" CW complex structure on $S^1 \times S^1$, $\gamma$ becomes the identity map on the non-collapsed $2$-cell.
