Is it true that:
For a map $f:X\rightarrow Y$, between two topological spaces. If the image of every convergent sequence in $X$ is also convergent in $Y$. Then $f$ is continuous.
If it is true, how to prove it? Or if it is false, what is the counter-example?
I guess it is false, because it is usually insufficient to characterize topological space with sequences. But I can't construct a counter-example. So I ask for help here.
Thanks for all the answers. Using nets or filters to characterize convergence seems to be a big topic such that I will spend some more time to digest. Before that, I seem to find an easy counter-example by myself.
Let $X=\{\{a\},\{a,b\},\emptyset\}$, every sequence in $X$ converges. The function $f$ from $X$ to $Y=\{\{f(a),f(b)\},\{f(b)\},\emptyset\}$. Then the image of every convergent sequence in $X$ is convergent in $Y$ but $f^{-1}(\{f(b)\})=\{b\}$ is not open in $X$, so $f$ is not continuous.
Best Answer
Construct a space which has a point which is not the limit of any sequence of points different from it, but which can be reached by a net (in other words, which is not isolated)
For example, let $\Omega$ be the smallest uncountable ordinal, let $X=\Omega+1$, and put the order topology on $X$. The biggest element $p$ in $X$ is such a point. Let $f:X\to X$ coincide with the identity map on $\Omega$ but with $f(p)=0$.