I'm trying to get my head around covering maps.
I've seen a question that asks to show there are covering maps $p$ from the torus $T$ onto $T$ with degree equal to any positive integer.
I'm not sure how to show this. I know there are covering maps from $S^1$ to $S^1$ with degree equal to any positive integer $n$ (by taking the $n$th power map), so I was wondering about mapping both the $x$ and $y$ coordinate of the torus to their $n$th power. However, I'm not sure if this has degree $n$ or degree $n^2$, and I'm not sure how to prove this is a covering map.
Any help much appreciated! Thank you.
First, do you know that the product of two covering maps $E\to B, F\to C$ (so $E\times F\to B\times C$) is also a covering map ?
If you don't already know this, you should definitely try and prove it, it's a good exercise !
With this in mind, it follows that for any $n,m\geq 1$, the map $p_{n,m} : S^1\times S^1\to S^1\times S^1$ given by $(z,w)\mapsto (z^n,w^m)$ is a covering map.
ADDED : Below, I used a notion of degree that is not the one you meant (they coincide for covering maps of manifolds).
For the notion you meant (the number of sheets of the covering, that is, the cardinality of each fiber), there is an easy proof which was essentially given by Paul Frost in the comments : $p_{n,m}^{-1}(x,y) = p_n^{-1}(x)\times p_m^{-1}(y)$ has cardinality $nm$.
Below was the old argument for the degree (using homological degree) :
Now, about its degree : since $n,m$ can be different, you see that your first guess (that the degree of $p_{n,n}$ be $n$) isn't quite reasonable. Your second guess, however, that it be $n^2$ (and more generally, that the degree of $p_{n,m}$ be $nm$) is way more reasonable, and in fact is what happens. And then you can clearly see how to get degree $n$ (for instance with $m=1$).
How to prove this ? Well it sort of depends on what you know. If you know the Künneth formula, then this will follow from it + the remark that $z\mapsto z^n, S^1\to S^1$ has degree $n$. Can you see why ?
If you don't know the Künneth formula, but do know about cohomology (or like it better), then you can do it using that as well : indeed by the universal coefficients theorem, the degree of the map can also be detected cohomologically, so looking at $H^2(\mathbb T)\to H^2(\mathbb T)$. But then $H^*(\mathbb T) = \mathbb Z[x,y]/(x^2,y^2)$ with $x,y$ in degree $1$ and anticommuting, so $H^2(\mathbb T) = \mathbb Z \cdot xy$, so you can get the degree from looking at where $x,y$ are sent. But for those you can look at $\pi_1$, and then it's clear.
If you don't know the Künneth formula, and don't know about the cohomology ring, there's probably other ways to do it, but probably with some trick. For instance use that $p_{n,m}$ is actually the composite of $p_{n,1}$ and $p_{1,m}$, so you can restrict yourself to either of those, and by symmetry to, say, $p_{n,1}$, and then you can probably use the local degree formula (but you'd have to know that).