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Let $X$ and $Y$ be two topological spaces and $f:X\to Y$ be a surjective map. If $intf(A)\subset f(intA)$ for any subset $A\subset X$, then $f$ is continuous. Here $int A$ means the interior of $A$

Can someone prove the claim or give some hints on it? Many Thanks!

The following solution is not true: Let $V\subset Y$ is open, noting that $V=f(f^{-1}(V))$, we have $$V=intV=intf(f^{-1}(V))\subset f(int(f^{-1}(V))),$$ which implies that $f^{-1}(V)\subset int f^{-1}(V)$. So, $f^{-1}(V)$ is open and $f$ is continuous. The problem is: for a set $B\subset X$, $f^{-1}(f(B))=B$ is not necessarily true since we don't know whether $f$ is injective.

ljjpfx
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  • This should follow directly from your hypothesis and the definition of continuity. – Andrew Mar 07 '23 at 23:11
  • What did you try? https://math.stackexchange.com/help/how-to-ask – Anne Bauval Mar 07 '23 at 23:12
  • I just found the answer here. https://math.stackexchange.com/questions/2384799/let-fx-to-y-be-a-surjective-map-such-that-textintfa-subset-f-textin?rq=1 – ljjpfx Mar 08 '23 at 01:32

1 Answers1

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This theorem is wrt surjective continuous maps. The idea is that f preserves interiors, where the interior of the image of a set is contained in the image of the interior of that set.

We prove this:

Assume intf(A)⊂f(intA) for all A⊂X. We want to show f is continuous. So take any open U⊂Y. We need to show f^-1(U) is open in X. But f^-1(U) = f^-1(intf(f^-1(U))) ⊂ int(f^-1(U)) by the assumption So f^-1(U) is open, and f is continuous.

Key is using the assumption on f to show that f^-1 preserves open sets. Hence, continuity follows easily.

In general, any time you have a map that "preserves some topological property" in this way, it's a good sign that continuity will follow. Preserving open sets, closed sets, interiors, etc. are common ways this happens.

Anon Imus
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  • @ Anon Imus Tnaks. How to get $ f^-1(intf(f^-1(U))) ⊂ int(f^-1(U))$? Would you like to present the details? Many thanks. – ljjpfx Mar 08 '23 at 01:18
  • Sure, here are the details:

    f^-1(intf(f^-1(U))) = f^-1(f(int(f^-1(U)))) by definition of intf ⊂ int(f^-1(U)) by assumption (intf(A) ⊂ f(intA) for all A)

    So we have: f^-1(U) = f^-1(intf(f^-1(U))) ⊂ int(f^-1(U))

    Which means f^-1(U) is open, and f is continuous.

    The key is just using the assumption that f preserves interiors (intf(A) ⊂ f(intA)) to relate f^-1(intf(f^-1(U))) and int(f^-1(U)), then the fact that the interior of a set contains the set gives us what we need.

     – Anon Imus Mar 08 '23 at 01:24