Proposition 10

If $\mathfrak{B}$ is an ultrafilter base on a set $X$ and if $f$ is a mapping of $X$ into a set $X'$, than $f(\mathfrak{B})$ is an ultrafilter base on $X'$.

Let $M'$ be a subset of $X'$. If $\overline{f}^{-1}(M')$ contains a set $M$ of $\mathfrak{B}$, then $M'$ contains $f(M)$; if not, then $\mathscr{C}_{\overline{f}}^{-1}(M') = \overline{f}^{-1}\mathscr{C}(M')$ contains a set $N$ of $\mathfrak{B}$ (no. 4, Proposition 5) and therefore $\mathscr{C}M'$ contains $f(N)$. Hence the result follows from Proposition 6 of no. 4.

Consider in particular the case where $f$ is the canonical injection $A \to X$ of a subset $A$ of a set $X$. If $\mathfrak{B}$ is a filter base on $A$ then $f(\mathfrak{B})$ is a filter base on $X$. The filter $\mathfrak{F}$ on $X$ generated by $f(\mathfrak{B})$ is called the filter generated by $\mathfrak{B}$ when $\mathfrak{B}$ is considered as a filter base on $X$. If $\mathfrak{B}$ is an ultrafilter base on $A$ it is also an ultrafilter base on $X$ by reason of Proposition 10.

Let us next examine whether the inverse image of a filter base is a filter base. Let $\mathfrak{B}'$ be a filter base on a set $X'$, and let $f$ be a mapping of a set $X$ into $X'$; then $\overline{f}^{-1}(\mathfrak{B}')$ is a filter base on $X$ if and only if $\overline{f}^{-1}(M') \neq \varnothing$ for each $M' \in \mathfrak{B}'$. This is an immediate consequence of the relation $\overline{f}^{-1}(M' \cap N') = \overline{f}^{-1}(M') \cap \overline{f}^{-1}(N')$ and of Definition 3 of no. 3. This condition can also be expressed by saying that every set of $\mathfrak{B}'$ meets $f(X)$ [or that the trace of $\mathfrak{B}'$ on $f(X)$ is a filter base]. If this condition is satisfied, then $f(\overline{f}^{-1}(\mathfrak{B}'))$ is a base of a filter finer than the filter of base $\mathfrak{B}$.

If $\mathfrak{B}$ is a filter base on $X$ it is clear that the above condition is satisfied by $\mathfrak{B}' = f(\mathfrak{B}); \; \overline{f}^{-1}(f(\mathfrak{B}))$ is then a base of a filter coarser than the filter of base $\mathfrak{B}$.

Let $A$ be a subset of a set $X$, $\varphi$ the canonical injection $A \to X$; if $\mathfrak{B}$ is a filter base on $X$ then $\overline{\varphi}^{-1}(\mathfrak{B})$ is the same as $\mathfrak{B}_A$. If we express this as a filter base of $A$ by means of the above condition, we recover part of Proposition 8 of no. 5.