I recently read that, given topological spaces $S,T$ and a map $f:S\rightarrow T$, for $f$ to be open it is sufficient to show that for a certain subbasis $C$ of $T$ and all (open) sets $A\in C$ holds $f(A)$ is open. (The converse holds by definition.)

This fact was stated as obvious, however I'm struggling with following it through. Before I prove the statement for any open set, I start with the open sets $A\in B$, where $B$ is the basis of $T$ obtained from $C$ (i.e. $B$ contains all finite intersections of Elements of $C$).

What I know:

  • $\exists A_1,\ldots,A_n\in C: A = \bigcap A_i$
  • since $A_i$ is open and in $C$, $f(A_i)$ is open by assumption for all $i=1,\ldots,n$
  • therefore $\bigcap f(A_i)$ is open

What I want to show:

  • $f(A)$ is open

If I had $f(A)=\bigcap f(A_i)$ I would be finished, but this doesn't hold generally. I don't know how to continue from here.

Getting from $C$ to any open set is easy, because for $A$ open exist index set $I$ and sets $A_i\in C$ with $i\in I$ such that $A=\bigcup A_i$, furthermore $f(A)=\bigcup f(A_i)$ always holds.

  • $\begingroup$ Your question says "subbasis" but your title says "prebasis". Which do you mean, and what is a "prebasis"? $\endgroup$ – bof Apr 24 '18 at 22:06
  • $\begingroup$ @bof Different literature, different authors. I corrected the title, coherent notation is better. I mean what I described in the question (the finite intersections) $\endgroup$ – SK19 Apr 24 '18 at 22:47

You cannot finish because the result is false: there are maps $f: X \to Y$ and a subbase $\mathcal{S}$ of $X$ such that $f[O]$ is open in $Y$ for all members $O$ of $\mathcal{S}$ but $f$ is not an open map.

A simple example: $X = \{0,1,2\}$ and $\mathcal{S} = \{\{0,1\}, \{1,2\}\}$. $Y = \{0,1\}$ in the indiscrete (trivial) topology and $f(0) = 0, f(1)=1 ,f(2)=0$. Then $f[O] = Y$ for all $O \in \mathcal{S}$ but $\{1\}$ is open in $X$ (as the intersection of two subbase elements) and its image is not open. We could add $\{0\}$ as an open set of $Y$, and both spaces would be $T_0$ (bit nicer).

It does hold for bases of $X$ instead of subbases as $f[\cup_i A_i] = \cup_i f[A_i]$ for all families of subsets of $X$ and any function $f$ on $X$.

  • $\begingroup$ Well, that explains it. The idea came from a German textbook where they wanted to show that the projections of product topologies are open and started their proof with "It's enough to show that the sets of a subbasis of the product topology have open images [under $pr_j$]" (transl.) They didn't indicate that this only works there or doesn't work in other cases. $\endgroup$ – SK19 Apr 24 '18 at 23:00
  • $\begingroup$ @SK19 for the product the standard base will do too. The projections of those are some $O_i$ or $X_i$, so always open. $\endgroup$ – Henno Brandsma Apr 25 '18 at 4:03

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