Remark

When $Y$ is Hausdorff, condition d) of Theorem 1 is equivalent to the following:

d') If $\mathfrak{U}$ is an ultrafilter on $X$ such that $f(\mathfrak{U})$ is a convergent filter base, then $\mathfrak{U}$ is convergent.

For if $\mathfrak{U}$ converges to $x$ and $f(\mathfrak{U})$ converges to $y$, then the uniqueness of the limit in $Y$ and the continuity of $f$ show that we must have $y = f(x)$. Likewise, Y being Hausdorff, condition c) of Theorem 1 is equivalent to:

c') If $\mathfrak{F}$ is a filter on X such that $f(\mathfrak{F})$ has a cluster point, then $\mathfrak{F}$ has a cluster point.

For c) $\Longrightarrow$ c') $\Longrightarrow$ d') $\Longrightarrow$ d) $\Longrightarrow$ c).

On the other hand, if Y is not Hausdorff, then d') no longer implies d); for example, take X to consist of one point and Y to consist of two points, with the coarsest topology.