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It is a theorem that if $f:X \to Y$ is a continuous bijection, $X$ is compact, then $g = f^{-1}$ is continuous. My professor asked us to find a counterexample to

If $f:X \to Y$ is continuous, $Y$ is compact, then $g = f^{-1}$ is continuous.

I do not like my counterexample so much because it uses the discrete metric. Are there other counterexamples?

My Counterexample: Let $X = [0, 1]$ with the Discrete metric, $Y = [0, 1]$ with the Euclidean metric, and let $f$ be the identity function.

Arturo Magidin
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Ovi
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  • @MariosGretsas Ah sorry I forgot to add that. I don't like my counterexample so much because it uses the discrete metric. Do you know of other counterexamples? – Ovi Dec 09 '19 at 18:33
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    The smallest example seems to be the identity $f:X\to Y$ with $X=Y={1,2}$, but $X$ with the topology ${\emptyset, {1}, X}$ and $Y$ with the topology ${\emptyset, Y}$. – topeik Dec 09 '19 at 18:36
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    @Ovi why not?? you counterexample is fine..if the inverse function was continuous then $f$ would be a homeomorphism which is not the case here, since the metric spaces in your example are not homeomorphic..one is compact and the other is not. – Marios Gretsas Dec 09 '19 at 18:38
  • @MariosGretsas It's just that if I can, I like to find counterexamples in more natural settings, such as functions from $\mathbb{R} \to \mathbb{R}$. – Ovi Dec 09 '19 at 18:40
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    If $f$ is a bijection, you can always identify $X$ and $Y$ as sets. Then $f$ being continuous corresponds to the $X$-topology being finer than the $Y$-topology. The inverse not being continuous then means it's strictly finer. Thus all examples are akin to yours, but of course one doesn't need to go the whole way to use the finest topology of all on $X$. Take $X = Y = [0,1]$, the standard topology on $Y$, the lower limit topology on $X$ for example. – Daniel Fischer Dec 09 '19 at 18:52
  • @DanielFischer Since the OP used the metric-spaces tag, I suspect that he or she is not interested on that example. – José Carlos Santos Dec 09 '19 at 18:54
  • Good point, @JoséCarlosSantos. Well, let's take $X = [0,1] \cup \mathopen{]} 2,3]$ and $Y = [0,2]$, both with the Euclidean metric. – Daniel Fischer Dec 09 '19 at 18:55
  • That is true, unfortunately I don't know any topology. – Ovi Dec 09 '19 at 18:56
  • @Ovi: given that, I've removed the general-topology tag. – Arturo Magidin Dec 09 '19 at 19:11
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    See my answer to a similar question for a simple counter-example. – TonyK Dec 09 '19 at 19:17
  • @ArturoMagidin Ah ok. I included it only because I thought topological people might be interested in answering this question. – Ovi Dec 09 '19 at 19:27
  • @Ovi: In that case, and given that you don't know topology, you should have made clear in your question what kind of answer you were looking for. Otherwise, you were bound to get (and did get) answers that ignored the metric and gave you topological examples that cannot be derived from a metric, and which are likely useless to you. – Arturo Magidin Dec 09 '19 at 19:29

2 Answers2

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Your example is fine, and quite simple and to the point.

Henno Brandsma
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Your example is fine. Another example is to take $X=[0,2\pi)$ with the usual metric, and let $Y=S^1 = \{z\in\mathbb{C}\mid |z|=1\}$, the unit circle on the complex plane, with the usual metric (so the distance is measured on the plane, not along the circle).

Let $f\colon X\to Y$ be given by $f(x) =e^{ix} = \cos(x) + i\sin(x)$.

Now, $Y$ is closed and bounded, so it is compact. The function $f$ is a continuous bijection. But $f^{-1}$ is not continuous; if it were, then $f$ would be a homeomorphism, but removing a single point from $X$ (other than $0$) will disconnect $X$, and removing a single point from $Y$ does not. Alternatively, note that you cannot find a neighborhood of $1\in Y$ whose image lies inside the neighborhood $[0,\frac{1}{2})$ of $f^{-1}(1)=0$, since the image will contain points arbitrarily close to $2\pi$.

Arturo Magidin
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