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I know if a function $f: X \to Y$ is bijective, it is open iff it is closed as both are equivalent to $f^{-1}$ being continuous.

Now I wonder if a function $f$ on two topological spaces is open and closed, must it be bijective? Furthermore, can it be neither injective nor surjective?

I suspect that there is some easy counterexample, but I cannot think of any.

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    $\begingroup$ Your $f$ is surjective $\endgroup$
    – FShrike
    Commented Jul 5, 2023 at 17:15
  • $\begingroup$ @FShrike Yeah I'm crazy, give a me a few minutes to fix $\endgroup$ Commented Jul 5, 2023 at 17:16
  • $\begingroup$ Actually I have not thought about topology for sometime, let me just edit it to bijective, and edit again if I thought about something. I'm not so sure how to temporarily delete a post and put it back $\endgroup$ Commented Jul 5, 2023 at 17:18
  • $\begingroup$ @TheSilverDoe I don't think this work, $\mathbb{R}$ is closed, but (-pi/2, pi/2) is open $\endgroup$ Commented Jul 5, 2023 at 17:24
  • $\begingroup$ @wsz_fantasy Yes of course. $\endgroup$ Commented Jul 5, 2023 at 17:26

3 Answers 3

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Let $X$ be any topological space with at least two points, and let $Y$ be any topological space with at least two points and an isolated point at 0. (E.g. $X=Y=\{0,1\}$ with the discrete topology.) Then $f:X\to Y$ defined by $f(x)=0$ is open, closed, not injective, and not surjective.

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Take any projection $\pi:X\times Y\to Y$ where $X$ is compact. It's both open (projections are open) and closed (since $X$ is compact; this is consequence of tube lemma).

If $h:Y\to Z$ is an embedding of $Y$ as a clopen subspace of $Z$, say $Z = Y\sqcup Y$ is disjoint union of two copies of $Y$, then $g = h\circ\pi$ is continuous, closed, open, not surjective, and not injective (for $|X| \geq 2$ and $Y\neq \emptyset$)

If you want an easier proof that $\pi$ is closed, you might take $X = Y$ to be compact Hausdorff. Then image of any closed $A\subseteq X^2$ is compact, hence closed since $X$ is Hausdorff.

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  • $\begingroup$ Why was I downvoted? $\endgroup$
    – Jakobian
    Commented Jul 5, 2023 at 17:29
  • $\begingroup$ It is not me, but can you elaborate with more details on why it is closed? $\endgroup$ Commented Jul 5, 2023 at 17:35
  • $\begingroup$ @wsz_fantasy this a standard fact from topology, consequence of tube lemma. See here. $\endgroup$
    – Jakobian
    Commented Jul 5, 2023 at 17:37
  • $\begingroup$ @Jackobian Thank you. I think it works. I'm not sure about the downvote, but I have given you an upvote. Would it be possible to have an example that is not surjective at the same time? $\endgroup$ Commented Jul 5, 2023 at 17:44
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    $\begingroup$ Also note that its image actually needs to be clopen, so all non-surjective maps like that are obtained from surjective ones in this way. $\endgroup$
    – Jakobian
    Commented Jul 5, 2023 at 17:57
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If we take $f: \mathbb{R} \mapsto \mathbb{R'}$ where $\mathbb{R}$ and $\mathbb{R'}$ are the real lines with indiscrete and discrete topologies respectively. We define the function as $f(x)=2 \forall x\in \mathbb{R}$, then $f$ is both open and closed but it is neither injective nor subjective.

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