Images vs Preimages for the (neighbourhood) filter. Let $f:X\to Y$ is an arbitrary function, $\mathcal{F}$ is a filter, and $X,Y$ are topological spaces.

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*Is it true (i think no) that
$$\{f[B]|B \in \mathcal{F}\}=\{U\subseteq Y | f^{-1}[U]\in\mathcal{F}\}$$
Assume that $f[B_1]=f[B_2]$, $U:=f[B_1]$. Then $f^{-1}[U]=B_1\cup B_2$. So for counterexample we need a filter $\mathcal{F}$ such that $B_1\in\mathcal{F}$ and $B_2\in\mathcal{F}$, but not $(B_1\cup B_2)\in\mathcal{F}$. So let's check the $Q := \{\{0\},\{0,1\},\{0,2\},\{0,3\},\{0,1,2,3\}\}$. It kind of has such property, but if $\mathcal{F}:=Q$ then it is not a  filter since it is not upward closed in $\mathcal{P}\{0,1,2,3\}$. What would be a good counterexample for the equality in question?


*Is it true if we use a neighbourhood filter in particular?
$$\mathcal{F}=\mathcal{N}(x)$$
This question is motivated by differences in definitions of "image filter" of ProofWiki and of Dugundji:
Dugundji 1966, pp. 215–221.
https://proofwiki.org/wiki/Definition:Image_Filter
 A: You seem to be confused about some things. In particular, I think you are not asking the right question.
First of, any definition of a filter I'm aware of, and the one on ProofWiki which you link, demands filters to be upwards closed. As a consequence, if $B_{1}\in\mathcal{F}$ then necessarily $B_{1}\cup B_{2}\in\mathcal{F}$ as $B_{1}\subseteq B_{1}\cup B_{2}$.
Secondly, a counterexample to 1. can be readily cooked up. Let $f$ be the function that maps all of $X$ to a fixed $y$ in $Y$. Then $\left\{f(B)\mid B\in\mathcal{F}\right\}=\left\{\left\{y\right\}\right\}$ but $f^{-1}(U)\in\mathcal{F}$ for any set containing $y$. Provided $Y$ contains at least two elements, this contradicts the claim. Note that our filter is completely arbitrary, so it also works for the neighbourhood filter. The moral of the story is that allowing arbitrary functions is almost always too ambitious.
The point is that Dugundji does not talk about filters here, but about filter bases. The two sets you give need not be equal, but the filters they generate are. The one on the right just happens to already be a filter.
