Here's a problem I've had a hard time with
If $f: M\rightarrow N$ is a cover map and $M$ is a m-manifold, will $N$ also be a m-manifold? A manifold is a space locally Euclidean space that is Hausdorff and Second Countable.
What if $N$ were a n-manifold, will $M$ also be a $n$-manifold?
Since I'm very new to topology, a heuristic explanation would be enough. I have some idea as to what the answers would be but I am not completely sure.
Suppose $f\colon M\to N$ is a covering map. If $N$ is a (topological) $n$-manifold, then $M$ is too. The converse, however, is true only with the additional assumption that $N$ is Hausdorff.
Here's a counterexample in which $N$ is not Hausdorff. Let $M = \mathbb R^2\smallsetminus \{(0,0)\}$, and let $N$ be the quotient space of $M$ by the equivalence relation generated by $$ (x,y) \sim (2^n x, 2^{-n}y), \qquad n\in\mathbb Z. $$ Then the quotient map $f\colon M\to N$ is a covering map, but $N$ is not Hausdorff, so it's not a manifold.
To see that $N$ is not Hausdorff, it's useful to look at the equivalence classes of $(1,0)$ and $(0,1)$.