What is a good criterion to quickly show a map is or is not closed? Definition: Let $X$ and $Y$ be topological spaces.  A function $f: X \rightarrow Y$ is closed (respectively, open) if, whenever $A \subseteq X$ is closed (resp. open) in $X$, then $f(A)$ is closed (resp. open) in $Y$.
I'm looking at the logarithm map $S^1 \rightarrow [0, 1)$ defined by $e^{2 \pi i t} \mapsto t$, which I think is a good example of a function that's open and closed but not continuous.
While looking at it, I realized that I don't really know of a quick way to show a map is closed.  (This is in comparison to being open, where you can just look at the image of a basic open set.)
With maps like the one given above, there are a couple cheesy ways I know:
(i) If it's bijective, being open and closed are equivalent.
(ii) If the function is continuous and the domain is compact (and the codomain is Hausdorff), then the map is automatically closed.
Is there anything more general?  Maybe a criterion in terms of nets?
 A: One classical one which is often useful, a sort of inverse continuity:
$f: X \rightarrow Y$ is closed iff for every $y \in Y$ and every open subset $O$ of $X$ such that $f^{-1}[\{y\}] \subseteq O$, there exists an open neighbourhood $U$ of $y$ in $Y$ such that $f^{-1}[U] \subseteq O$.
Proof sketch: if $f$ is closed, and $y, O$ are as stated, define $U = Y \setminus f[X \setminus O]$ which is open as $f$ is closed. Pick $x \in f^{-1}[U]$. If $x$ were not in $O$, $f(x)$ would be in $f[X \setminus O]$, but $f(x) \in U$, which is its complement. So $x \in O$ as required, and $f^{-1}[U] \subseteq O$.
If the criterion holds, let $C \subseteq X$ be closed. Suppose $y \notin f[C]$.
Then $f^{-1}[\{y\}] \subseteq X \setminus C$ (no point that maps to $y$ can be in $C$), and applying the right hand side criterion to this $y$ and $O = X \setminus C$ which is open, we find an open $U \subseteq Y$ with $y \in U$ such that $f^{-1}[U] \subseteq X \setminus C$, which says that no point that maps into $U$ can lie in $C$, or equivalently $U \cap f[C] = \emptyset$. This shows that no point outside $f[C]$ lies in its closure, so $f[C]$ is closed, and so $f$ is a closed map.
Another simple one in terms of closure:
$f$ is closed iff for all subsets $A \subseteq X$, $\overline{f[A]} \subseteq f[\overline{A}]$.
If $f$ is closed, $f[A] \subseteq f[\overline{A}]$ and the right hand side is closed, so the inclusion holds.
If the inclusion holds for all subsets $A$, pick $C$ closed. Then $f[C] \subseteq \overline{f[C]} \subseteq f[\overline{C}] = f[C]$. So $f[C] = \overline{f[C]}$ and $f[C]$ is closed.
Fun fact: the reverse inclusion holds for all $A \subseteq X$ iff $A$ is continuous. So again it's a dual to continuity in a way.
