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Let $f$ be a map between topological spaces $f : X \to Y$ such that for every $A \subset X$ we have that $f(\operatorname{int} A) = \operatorname{int} f(A)$. Prove $f$ is continuous.

I think I can prove the converse isn't true, any continuous non-open map should do, take $f : \mathbb{R} \to \mathbb{R} : x \mapsto x^2$ with the standard topology, and $A = \mathbb{R} = \operatorname{int}A$. Then $f(\operatorname{int}A) = [0, +\infty) \neq (0, +\infty) = \operatorname{int}f(A)$

I have no clue how to prove this. I have tried using the fact that $f$ is continuous if $f(\overline{B}) \subset \overline{f(B)}$ for all $B$, but I can't seem to construct the right $A$ that implies this.

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    $\begingroup$ Concerning the converse: The projection $p : [0,1] \times [0,1] \to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(\operatorname{int} A) = \operatorname{int} p(A)$ (consider $A$ = diagonal). $\endgroup$
    – Paul Frost
    Jan 27, 2019 at 17:17
  • $\begingroup$ What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples? $\endgroup$
    – Paul Frost
    Jan 27, 2019 at 23:24
  • $\begingroup$ I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing. $\endgroup$
    – Kasper
    Jan 28, 2019 at 0:07
  • $\begingroup$ Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it. $\endgroup$
    – Paul Frost
    Jan 28, 2019 at 16:01
  • $\begingroup$ I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks. $\endgroup$
    – Kasper
    Jan 29, 2019 at 1:03

2 Answers 2

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This is only a partial answer. Obviously $f$ is open. Concerning continuity we may w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X \stackrel{f}{\rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(\operatorname{int} A) = \operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B \subset f(X)$ we have $\operatorname{int}_Y B = \operatorname{int}_{f(X)} B$).

Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V \subset Y$ be open. Then $f(\operatorname{int} ( f^{-1}(V)) = \operatorname{int} f(f^{-1}(V)) = \operatorname{int} V = V = f(f^{-1}(V))$, hence $\operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.

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From a rough sketch given by the TA of the course, I was able to formulate a proof.

Consider an open $B \subset Y$. Let $A = f^{-1}(B)$. Let $C = A \setminus \operatorname{int} A$. The goal is to prove $C$ is empty.

Consider now $G = C \cup (A \setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(\operatorname{int} G) = \operatorname{int}(f(G)) = \operatorname{int} B = B$

Now $\operatorname{int}G =\operatorname{int} C \cup \operatorname{int}(A \setminus f^{-1}(f(C))) = \operatorname{int}(A \setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(\operatorname{int} G) = f(\operatorname{int} (A \setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A \setminus f^{-1}(f(C))) = f(A) \setminus f(C) = B \setminus f(C)$

So $B = B \setminus f(C)$, which implies that $f(C) = \varnothing$ and thus $C = \varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.

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  • $\begingroup$ Here are two problems in the proof. (1) In general $f(A) \subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B \cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B \subset f(X)$. In that case $f(A) = B$. (2) We only have $\operatorname{int} (M \cup N) \supset \operatorname{int} M \cup \operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $\operatorname{int}G = \operatorname{int}(A \setminus f^{-1}(f(C)))$. $\endgroup$
    – Paul Frost
    Jan 29, 2019 at 10:52
  • $\begingroup$ Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A \setminus \operatorname{int} A \subset \overline{A} \setminus \operatorname{int} A = \partial A$, and $A \setminus f^{-1}(f(C)) \subset A \setminus C = A \setminus (A \setminus \operatorname{int}A) = \operatorname{int} A$. And the boundary and interior of a set are always separate. $\endgroup$
    – Kasper
    Jan 29, 2019 at 20:18
  • $\begingroup$ The usual definition that sets $M,N$ are separated is that $\overline{M} \cap N = M \cap \overline{N} = \emptyset$. But $\partial A$ and $\operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G \subset A$. Hence $\operatorname{int} G \subset \operatorname{int} A$ so that $\operatorname{int} G \cap \partial A = \emptyset$. Since $C \subset \partial A$, we get $\operatorname{int} G \subset A \setminus f^{-1}(f(C))$ whence $\operatorname{int} G \subset \operatorname{int} (A \setminus f^{-1}(f(C))) \subset \operatorname{int} G$. $\endgroup$
    – Paul Frost
    Jan 30, 2019 at 0:50

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