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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.

wsz_fantasy
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3 Answers3

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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.

Steven Clontz
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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.

Jakobian
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  • Why was I downvoted? – Jakobian Jul 05 '23 at 17:29
  • It is not me, but can you elaborate with more details on why it is closed? – wsz_fantasy Jul 05 '23 at 17:35
  • @wsz_fantasy this a standard fact from topology, consequence of tube lemma. See here. – Jakobian Jul 05 '23 at 17:37
  • @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? – wsz_fantasy Jul 05 '23 at 17:44
  • Yes, you just have to embedd $Y$ as a proper clopen subset of some $Z$. For example, take $Z$ to be disjoint union of two copies of $Y$ i.e. $Z = {0}\times Y \cup {1}\times Y$ where $U\subseteq Z$ is open iff $U\cap {i}\times Y$ are open for $i = 0, 1$, where ${i}\times Y$ are treated as copies of $Y$ (i.e. $(i, y)\mapsto y$ is a homeomorphism). The map $g:X\times Y\to Y\cong {0}\times Y\to Z$ (composition of $\pi$, homeomorphism of $Y$, and an embedding) will still be open and closed, but not surjective. – Jakobian Jul 05 '23 at 17:48
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    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. – Jakobian Jul 05 '23 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.

Arohan
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