Prove or disprove that if $f:X\to Y$ is surjective then $f(\mathcal F)$ is a(n) (ultra)filter for any (ultra)filter $\mathcal F$ in $X$.

Question

I know that if $f$ is a function from a set $X$ to a set $Y$ then for any filter $\mathcal F$ in $X$ the collection $$ f(\mathcal F):=\big\{V\in\mathcal P(Y):V=f[F]\,\text{with }F\in\mathcal F\big\} $$ is not generally a filter but it is always a filter base: however is it true that $f(\mathcal F)$ is a filter when $f$ is surjective? Indeed, if $\mathcal F$ is a nonempty collection of nonempty sets then $f(\mathcal F)$ is such. Moreover, if $V\in\mathcal P(Y)$ is such that $$ f[F]\subseteq V $$ for any $F\in\mathcal F$ then it is also such that $$ F\subseteq f^{-1}\big[f[F]\big]\subseteq f^{-1}[V] $$ so that $f^{-1}[V]$ is in $\mathcal F$ and thus by surjectivity $V$ is in $f(\mathcal F)$. Finally given $F_i\in\mathcal F$ for $i=1,2$ the inclusion $$ F_1\cap F_2\subseteq f^{-1}\big[f[F_1]\big]\cap f^{-1}\big[f[F_2]\big]=f^{-1}\big[f[F_1]\cap f[F_2]\big] $$ holds so that if $F_1\cap F_2$ lies in $\mathcal F$ then also $f^{-1}\big[f[F_1]\cap f[F_2]\big]$ does it and thus by surjectivity we conclude that $f[F_1]\cap f[F_2]$ lies in $f(\mathcal F)$. So $f(\mathcal F)$ is a nonempty collection of nonempty sets closed by finite intersection and by upward inclusion, that is $f(\mathcal F)$ is a filter in $Y$.

Now we remember (see here for details) that a collection $\mathcal U$ is an ultrafilter in a set $X$ if and only if it is a centered system of sets such that $Y\in\mathcal U$ or $X\setminus Y\in\mathcal U$ for any $Y\subset X$: so if $\mathcal U$ is an ultrafilter in $X$ then $f^{-1}[V]\in\mathcal U$ or $X\setminus f^{-1}[V]\in\mathcal U$ so that by the identity $$ X\setminus f^{-1}[V]=f^{-1}[Y]\setminus f^{-1}[V]=f^{-1}[Y\setminus V] $$ and by surjectivity of $f$ we conclude that $$ V\in f(\mathcal U)\quad\text{or}\quad Y\setminus V\in f(\mathcal U) $$ which proves that $f(\mathcal U)$ is an ultrafilter.

So I would like to know if what I observed is effectively true and thus if it is well proved: so could someone help me, please?

Answer 1

Your proof is correct.

The key observation is that if $f$ is surjective, then $G \in f(\mathcal F)$ iff $f^{-1}(G) \in \mathcal F$. In fact, if $G \in f(\mathcal F)$, then $G = f(F)$ for some $F \in \mathcal F$, thus $f^{-1}(G) = f^{-1}(f(F)) \supset F$ and therefore $f^{-1}(G) \in \mathcal F$ (this is true for any $f$). Conversely, if $f^{-1}(G) \in \mathcal F$, then $G = f(f^{-1}(G)) \in f(\mathcal F)$ by surjectivity.

1. $f(\mathcal F)$ is closed by by upward inclusion:
   Let $f(F) \subset G$. Then $F \subset f^{-1}(f(F)) \subset f^{-1}(G)$, thus $f^{-1}(G) \in \mathcal F$.

2. $f(\mathcal F)$ is closed by by finite intersection:
   We have $f(F_1 \cap F_2) \subset f(F_1) \cap f(F_2)$. Since $f(F_1 \cap F_2) \in f(\mathcal F)$, we see that $f(F_1) \cap f(F_2) \in f(\mathcal F)$.

3. If $\mathcal U$ is an ultrafilter, then so is $f(\mathcal U)$:
   Let $V \subset Y$. Then $f^{-1}(V) \subset X$, thus $f^{-1}(V) \in \mathcal U$ or $f^{-1}(Y \setminus V) = X \setminus f^{-1}(V) \in \mathcal U$ .