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When determining whether functions are continuous, it is ok to check whether the preimage of basis elements in the codomain are open in the domain? Is the same thing true for open and closed maps? Can we simply check whether they map basis elements in the domain, to open sets in the codomain?

Trying to prove it I realize I run into issues in that for any open set $U$ in the domain, unlike preimages which behave nicely on unions and intersections of sets if $U=\bigcup\limits_{\alpha \in J}B_\alpha$, where $B_\alpha$ are basis elements for the topology in which $U$ is open the $f(U) \neq \bigcup\limits_{\alpha \in J}f(B_\alpha)$ so I suppose this would not hold unless $f$ is continuous and bijective. Or would a different set of conditions allow this to hold?

Scott Frazier
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    Yes, you can check continuity by inspecting a basis of the codomain, and you can check openness by inspecting a basis of the domain. I don’t think you can check closedness using a basis, however. – Zhen Lin Apr 05 '21 at 04:43
  • @Zhenlin I am looking at a book on set theory and it says $f(A \cup B)=f(A) \cup f(B)$ so I am guessing that's why checking for openness suffices. However it does not look like checking openness can be done on subbasis elements since $f(A \cap B) \subset f(A) \cap f(B)$. But if $f$ is injective then equality holds. – Scott Frazier Apr 05 '21 at 04:50
  • I show in this answer that the checking on base elements does work. And $f[U]=\bigcup_{\alpha \in J} f[B_\alpha]$ trivially holds. Elementary set theory... – Henno Brandsma Apr 05 '21 at 08:25

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