Why is this a quotient map

Question

Is there a direct way to see that $p \times id : [0,1]^2 \rightarrow S^1 \times [0,1]$ is a quotient map with $(p \times id)(x,y) = (e^{ix},y)$? By direct way, I mean is there an obvious argument why this has to be a quotient map? I somehow feel that there is an obvious reason why this has to be one, but I don't see it.

Answer 1

There is indeed an obvious reason:

Continuous maps from quasicompact spaces to Hausdorff spaces are closed.

And continuous and closed surjective maps are quotient maps.

Answer 2

One can use either of the following three facts to show that $q\times 1_Y$ is closed.

Fact 1: If $q:X\to Z$ is a quotient map, and $Y$ is locally compact, then $q×1_Y:X×Y\to Z×Y$ is a quotient map.

In your case, $Y=[0,1]$ which is locally compact.

A map is called perfect if it's closed and all fibers are compact.

Fact 2: If $q:X\to Z$ is a surjective perfect map, then $q×1_Y$ is a closed surjection for any space $Y$, so it is a quotient map.

Note that a map from a compact to a Hausdorff space is always perfect. This also implies

Fact 3: If $q:X\to Z$ is any surjective map, $X$ is compact, $Z$ is Hausdorff, and $Y$ is compact Hausdorff, then $q×1_Y$ is a closed surjection, as $X\times Y$ is compact and $Z\times Y$ is Hausdorff.

This is of course the statement that you would most likely use in your situation, since all your spaces are compact Hausdorff. The other facts can be applied if some of your spaces are not.