Convergent Nets and Composite Functions Does anyone have any idea how to prove the following...?
Let $X, Y$ be topological spaces and let $g:X\rightarrow Y$ be a map. If for every directed set $I$ and convergent net $f:I\rightarrow X$ with limit $x$, the composite $g\circ f:I\rightarrow Y$ is a convergent net with limit $g(x)$, then $g:X\rightarrow Y$ is continuous at $x\in X$.
Many thanks!
 A: Let's recall what is means for $g$ to be continuous at $x$:
$$\forall V \text{ open in } Y: (g(x) \in V) \implies (\exists U \text{ open in X }: x \in U \text{ and } g[U] \subseteq V)$$
And consider the "canononical net" that converges to $x$:
Let $I = \mathcal{N}_x$, the set of all neighbourhoods of $x$, directed by 
$U \ge V$ iff $U \subseteq V$. Now for any $f: I \to X$ that obeys $\forall U \in I: f(U) \in U$, (so for every neighbourhood of $x$ we pick a point of that neighbourhood), $f \to x$ or $x \in \lim f$ (or whatever your notation for "$x$ is a limit of $f$" is). This is clear by construction: if $O$ is an open set containing $x$, then $O \in I$ so we choose this as $i_0$ and if $i \ge i_0$ we have that $f(i) \in i \subseteq i_0= O$, so $\forall i \ge i_0: f(i) \in O$ and as $O$ was arbitary we have convergence to $x$.
Now suppose that $g$ obeys the "net-composition property", and that $g$ is not continuous at $x$. This means (by negating the definition I gave at the outset) that there is some fixed $V$, open in $Y$ that contains $g(x)$ but such that no open neighbourhood $U$ of $x$ obeys $g[U] \subseteq V$. If we state this more "positively", for every neighbourhood $U$ of $x$, there is some $x_U \in U$ that "witnesses" that $g[U] \nsubseteq V$, or $g(x_U) \in g[U]$ (obviously) but $g(x_U) \notin V$. 
So this non-continuity assumption gives us a net $f: I \to X$ defined by $f(U) = x_U$ for any $U \in \mathcal{N}_x$, as defined above. Then $x \in \lim f$ as we saw (we picked $x_U \in U$ for all $U$) but $g(f(U)) \notin V$ for all $U$, and this immediately implies that $g(x)$  is not a limit of $g \circ f: I \to Y$: a whole neighbourhood of $g(x)$ contains not a single point from that net. This contradicts the "net-composition property", so our assumption that $g$ was not continuous at $x$ was false, hence $g$ is continuous at $x$ as required.
A: As an alternative to the standard proof that directly constructs a net to apply the hypothesis to, here's a proof that reuses the fact you've probably already seen: that the closure of $S \subseteq X$ is equal to the set of limits of nets with values in $S$.
So, in order to use this fact, we will use this characterization of continuity at a point: $g$ is continuous at $x$ if and only if whenever $S \subseteq X$ and $x \in \overline{S}$, then $g(x) \in \overline{g(S)}$.  (Outline of the $\Leftarrow$ direction which is the one we will use: suppose, for contradiction, that $V \subseteq Y$ is a neighborhood of $g(x)$, but $g^{-1}(V)$ is not a neighborhood of $x$.  Then that means that $x$ is in the closure of $X \setminus g^{-1}(V) = g^{-1}(Y \setminus V)$, but $g(x)$ is not in the closure of $Y \setminus V$, which implies that $g(x)$ is also not in the closure of $g(g^{-1}(Y \setminus V)) \subseteq Y \setminus V$, contradicting the hypothesis.)
So now, suppose we have the hypothesis that every net converging to $x$ is transformed by $g$ to a net converging to $g(x)$.  Then, suppose $S \subseteq X$ and $x \in \overline{S}$.  Then there exists a net $f : I \to X$ converging to $x$ such that $f(i) \in S$ for every $i \in I$.  By the hypothesis, then $g \circ f : I \to Y$ converges to $g(x)$; but we also have $(g \circ f)(i) = g(f(i)) \in g(S)$ for every $i \in I$, implying $g(x) \in \overline{S}$ as desired.

As yet another approach to the proof, you could look at proving: $N \subseteq X$ is a neighborhood of $x$ if and only if for all directed sets $I$ and nets $f : I \to X$, if $f$ converges to $x$, then $f(i)$ is eventually in $N$.  Once you have that, the proof would be easy: suppose $V$ is a neighborhood of $g(x)$; we want to prove that $g^{-1}(V)$ is a neighborhood of $x$.  Using the mentioned fact, suppose that $f : I \to X$ converges to $x$; then by the problem's hypothesis, $g \circ f : I \to Y$ converges to $g(y)$, so for large enough $i \in I$, $(g \circ f)(i) \in V$.  This implies that for large enough $i \in I$, $f(i) \in g^{-1}(V)$, which is what we needed to establish.
And the proof of the neighborhood-in-terms-of-nets fact would go similarly: the $\Rightarrow$ part is true almost directly from the definition of net convergence.  For the $\Leftarrow$ part, just use the fact that $N$ is a neighborhood of $x$ if and only if $x \in \operatorname{int}(N) = X \setminus \overline{X \setminus N}$, if and only if $x \notin \overline{X \setminus N}$.
