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I'm currently revisiting a proof of Tychonoff's theorem via ultrafilters. The definitions we were working with are as follows,

• 
  $mathcal{B}$ is a basis for a filter $mathcal{F}$ on a set $X$ if $mathcal{F} = {A : B subset A, text{for some $B in mathcal{B}$}}$.


• an ultrafilter on a set $X$ is a filter $mathcal{F}$ that is a maximal element of the set of filters on $X$, ordered by inclusion.

As an intermediate step, a lemma was left for the reader to prove:

Lemma. Let $f : X to Y$ be a function. Then,

• if $mathcal{B}$ is a basis for a filter on $X$, then $f(mathcal{B})$ is a basis for a filter on $Y$




• if $mathcal{B}$ is a basis for a filter on $Y$, then $f^{-1}(mathcal{B})$ is a basis for a filter on $X$, provided that $f^{-1}(B) neq emptyset$ for all $B in mathcal{B}$.




• if $mathcal{B}$ is a basis for an ultrafilter on $X$, then $f(mathcal{B})$ is a basis for an ultrafilter on $Y$

I have managed to prove the first two stamentents, but I am struggling with the last one. Certainly $f(mathcal{B})$ is a basis for a filter on $Y$, but why is this an ultrafilter?

I've attempted to assume the contrary and take a finer filter than the one generated by $f(mathcal{B})$, in order to take preimages and contradict that $mathcal{B}$ generates an ultrafilter, but I haven't been able to complete the proof.

For most purposes, the most useful characterization of ultrafilters is not as maximal filters but as filters with the property that for any $Asubseteq X$, either $A$ or $Xsetminus A$ is in the filter. (Prove this is equivalent to maximality, if you haven't seen this! The key point is that if neither $A$ nor $Xsetminus A$ is in the filter, you can add $A$ to get a larger filter.)

So, now suppose $mathcal{B}$ is a basis for an ultrafilter on $X$. This means that for any $Asubseteq X$, there exists $Bin mathcal{B}$ such that either $Bsubseteq A$ or $Bsubseteq Xsetminus A$. So now, to prove that $f(mathcal{B})$ is a basis for an ultrafilter, let $Asubseteq Y$ be arbitrary. We want to show there is $Binmathcal{B}$ such that either $f(B)subseteq A$ or $f(B)subseteq Ysetminus A$. Note that these inclusions are equivalent to $Bsubseteq f^{-1}(A)$ or $Bsubseteq Xsetminus f^{-1}(A)$. So, we know there exists such a $B$ because $mathcal{B}$ is a basis for an ultrafilter on $X$.

Assume $mathcal{B}$ is a base for an ultrafilter $mathcal{U}$.

By point 1 we know that $f[mathcal{B}]$ generates a filter too, call it $mathcal{F}$. Each $F in mathcal{F}$ contains some $f[B]$ for $B in mathcal{B}$ so $f^{-1}[F]$ is non-empty (it contains $B$) and so $f^{-1}[mathcal{F}]$ generates a filter on $X$ by point 2.

If $B in mathcal{B}$ then $f[B] in f[mathcal{B}]$, so $f[B] in mathcal{F}$ and as $B subseteq f^{-1}[f[B]]$, we have that $f^{-1}[f[B]] in mathcal{U}$, and as this also holds for all supersets of $f[B]$ we see that $f^{-1}[mathcal{F}] = mathcal{U}$. But this immediately implies that $mathcal{F}$ is maximal and so $f[mathcal{B}]$ is an ultrafilter base.