TITLE: $f$ brings convergent nets to convergent nets, is it continuous?
QUESTION [14 upvotes]: Let $f:(X,\mathcal T)\to (Y,\mathcal S)$ be a function between topological spaces. Let for any convergent net $(x_\alpha)$ in $X$, $(f(x_\alpha ))$ be convergent in $Y$. Is $f$ continuous?
(It seems to be true in completely regular spaces).

REPLY [16 votes]: The "correct" characterization of continuity of functions using nets is:
Proposition: Let $X$ and $Y$ be topological spaces.  For a map $f: X \rightarrow Y$, the following are equivalent:
(i) For every convergent net ${\bf x} \rightarrow x$ in $X$, $f({\bf x}) \rightarrow f(x)$ in $Y$.
(ii) $f$ is continuous.
This is Proposition 3.2 in these notes.  The proof is left as an exercise.  Here is how to prove (i) $\implies$ (ii): combine Proposition 3.1 -- The closure of a subset $S$ in a topological space is the set of limits of convergent nets with values in $S$ -- with Theorem 3 from these notes: a function $f: X \rightarrow Y$ is continuous iff for every subset $S$ of $X$, $f(\overline{S}) \subset \overline{f(S)}$.  (The proof of (ii) $\implies$ (i) is the same easy proof as for sequences.)
The OP's question is subtly different from this, since he says only that $f(\bf{x})$ is convergent, not that it converges to $f(x)$.  The very next result in my notes on convergence is Proposition 3.3: a space $X$ is Hausdorff iff every net in $X$ converges to at most one point.
So the answer to the OP's question is yes when $Y$ is Hausdorff.  (Added: to see this one implements a net-theoretic analogue of the construction in Stefan H.'s answer.  Let me know if more details on this are desired.)
One should be skeptical of the statement otherwise, and indeed Stefan H. has given a nice counterexample in his answer: presumably one of the simplest possible ones.  Note that a feature of his construction is to take a space $Y$ in which there is a point $y$ which admits no proper open neighborhood.  Any net in such a space must converge to $y$!