$f$ closed iff $y\in N$ and open $V\supset f^{-1}\left(\{y\}\right)$ exists $U$ open such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$ Prove that $f\colon M\to N$ (topological spaces) is closed if and only if for all $y\in N$ and all open sets $V\supset f^{-1}\left(\{y\}\right)$ in $M$ there exists an open set $U$ in $N$ containing $y$ such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$.
I can't prove this in any way. I tried for 3 days.
 A: I was able to obtain one direction:
Fix some $y\in N$. Note that $V\supseteq f^{-1}(U)\supseteq f^{-1}(\{y\})$ if and only if $$V^c\subseteq f^{-1}(U)^c= f^{-1}(U^c)\subseteq f^{-1}(\{y\})^c.$$
If $f$ is closed, then for any closed set $C\subseteq M$ disjoint from $f^{-1}(\{y\})$, we have that $D=f(C)$ is a closed set disjoint from $\{y\}$, and $$C\subseteq f^{-1}(f(C))=f^{-1}(D)\subseteq f^{-1}(\{y\})^c.$$ Then $V=C^c$ and $U=D^c$ are open subsets of $M$ and $N$ respectively satisfying the necessary property, and every open subset $V\supseteq f^{-1}(\{y\})$ of $M$ can of course be obtained as the complement of a closed subset $C$ of $M$ that is disjoint from $f^{-1}(\{y\})$.
A: $(\Rightarrow)$ Zev Chonoles has already given this direction. I give the other. 
$(\Leftarrow)$ Suppose $C \subset M$ is closed. Want to show $f(C)$ is closed in $N$. Suppose $y$ is a limit point of $f(C)$. First, you want to show that $f^{-1}(\{y\}) \neq \emptyset$. Suppose $f^{-1}(\{y\}) = \emptyset$. Then let $V = \emptyset$. Then for any $U \subset N$ containing $y$, since $y$ is a limit point of $f(C)$, this $U$ must contain a point of $C$. So $f^{-}(U) \neq \emptyset$. Contradiction. It is impossible that $V = \emptyset \supset f^{-1}(U) \supset f^{-1}(\{y\}) = \emptyset$. I have so far shown that $f^{-1}(\{y\}) \neq \emptyset$. 
Now I claim is that at least one point in $f^{-1}(y)$ is a limit point of $C$. Suppose not, then every $x \in f^{-1}(y)$ has neighborhood $B_x$ such that $B_x \cap C = \emptyset$. Let $V = \bigcup_{x \in f^{-1}(y)}B_x$. $V$ is a neighborhood of $f^{-1}(y)$ such that $V \cap C = \emptyset$. For every $U \subset N$ containing $y$, $U$ contains a point of $C$ since $y \in U$ is a limit point of $f(C)$. So $f^{-1}(U) \cap C \neq \emptyset$ so it is impossible that $V \supset f^{-1}(U) \supset f^{-1}(\{y\})$. So I have shown that at least one point $z \in f^{-1}(y)$ is a limit point of $C$. Since $C$ is closed, $z \in C$. So $f(z) = y$. So $y \in f(C)$. $f(C)$ contains all its limit points. So it is closed. $f$ is a closed map. 
A: There’s even a pointwise version of the result. I denote the topology of a space $X$ by $\tau(X)$.
For $y\in N$ let 
$$\begin{align*}
\mathscr{N}(y)&=\{U\in\tau(N):y\in U\},\\
\mathscr{B}_y&=\{f^{-1}[U]:y\in U\in\tau(N)\},\text{ and}\\
\mathscr{V}_y&=\{V\in\tau(M):f^{-1}[\{y\}]\subseteq V\}\;,
\end{align*}$$ 
and let $\mathscr{F}_y$ be the filter on $M$ generated by $\mathscr{B}_y$; the claim is that $f$ is closed iff for each $y\in N$, $\mathscr{V}_y\subseteq\mathscr{F}_y$.

Definition: The function $f$ is closed at $y\in N$ iff for each closed $K\subseteq M$, $y\in\operatorname{cl}f[K]$ iff $y\in f[K]$.

Proposition: Let $y\in N$; then $f$ is closed at $y$ iff $\mathscr{V}_y\subseteq\mathscr{F}_y$.

Proof: Suppose that $\mathscr{V}_y\nsubseteq\mathscr{F}_y$, and fix $V\in\mathscr{V}_y\setminus\mathscr{F}_y$. Let $K=M\setminus V$; $K$ is closed, and $K\cap f^{-1}[\{y\}]=\varnothing$, so $y\notin f[K]$. Since $V\notin\mathscr{F}_y$, for each $U\in\mathscr{N}(y)$ there is a point $x_U\in f^{-1}[U]\setminus V=K\cap f^{-1}[U]$; clearly $f(x_U)\in U\cap f[K]$, so $y\in\operatorname{cl}f[K]$, and therefore $f[K]$ is not closed.
Conversely, suppose that $y\in\operatorname{cl}f[K]\setminus f[K]$ for some closed $K\subseteq M$. Let $V=M\setminus K$; clearly $V\in\mathscr{V}_y$. Suppose that $U\in\mathscr{N}(y)$; $y\in\operatorname{cl}f[K]$, so $U\cap f[K]\ne\varnothing$, and therefore $$f^{-1}[U]\setminus V=f^{-1}[U]\cap K\ne\varnothing\;.$$ Thus, $V\notin\mathscr{F}_y$, and hence $\mathscr{V}_y\nsubseteq\mathscr{F}_y$. $\dashv$

A: I think that I have the other direction using the same idea that Zev Chonoles ♦:
Let $F\subset M$ a closed subset. Let some $y\in f[F]^c$ then 
$$f^{-1}[\{y\}]\subset f^{-1}[f[F]^c]=f^{-1}[f[F]]^c\subset F^c,$$
in particular $f^{-1}[\{y\}]\subset F^c$ where $F^c$ is open. Therefore exists an open $U$ containing $y$ such that 
$$f^{-1}[U]\subset F^c\implies F\subset f^{-1}[U^c]$$
therefore
$$f[F]\subset f[f^{-1}[U^c]]\subset U^c,$$
in particular
$$y\in U\subset f[F]^c.$$
Therefore $f[F]^c$ is open and $f[F]$ is closed.
