Injective map preserves ultrafilter (?) Proof This is a theorem in Willard's text, Theorem 12.14, p81, which states:

$f$ maps $X$ into $Y$ and $\mathcal{F}$ is an ultrafilter on $X$, then $f(\mathcal{F})$ is an ultrafilter on $Y$.
We define $f(\mathcal{F})$ to be filter with $\{ f(F) : F \in \mathcal {F} \}$ as filter base.

I just wanted to check if this is right, because my proof did not use injectivity rather surjectivity.
Proof: Suppose false: there is filter $\mathcal{G}$ such that $f(\mathcal{F}) \subseteq  \mathcal{G} $ and $G' \in \mathcal{G} \setminus f(\mathcal{F})$. Consider filter on $X$ with filter base $\{ f^{-1}(G) : G \in \mathcal{G} \}$, then this is a strictly finer filter of $\mathcal{F}$, if $f$ is onto, with $f^{-1}(G')$ nonempty.
 A: No, you don't need that $f$ is surjective, because $G'\cap f(X)$ cannot be empty, as $\cal G$ is a filter and $f(X)\in\cal G$. So you get the filter that you defined is indeed a filter extending $\cal F$. The rest of the proof is fine.
The point is that $A=f^{-1}(G\cap f(X))$ is the only subset satisfying $f(A)=G\cap f(X)$. Because $f$ is injective. So the above argument cannot fail.
A: If $A \subset Y$, consider $f^{-1}[A]$: if it is empty, then $f[X] \subset Y\setminus A$, and so $Y \setminus A \in f(\mathscr{F})$. If $f^{-1}[A] \in \mathscr{F}$, then $f[f^{-1}[A]] \subset A$ implies $A \in f(\mathscr{F})$. If on the other hand $X\setminus f^{-1}[A]\in \mathscr{F}$, then  $f[X \setminus f^{-1}[A]] \subset Y \setminus A$ (if $x \in X\setminus f^{-1}[A]$ , $f(x) \notin A$ by definition). The latter implies  that $Y \setminus A \in f(\mathscr{F})$. 
This uses the characterisation of an ultrafilter that for every set $A$ either $A$ or its complement is in it. I don't see any assumptions on $f$ that I need. Note that the first case is actually subsumed under the final case. So there are really two cases: $f^{-1}[A] \in \mathscr{F}$ or not.
