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If $f:M\to N$, $g:N\to P$ continuous and $g\circ f: M\to P$ is a homeomorphism. And $g$ is injective (or $f$ is surjective) then $g, f$ both are homeomorphisms.

I don't know how to prove it. I tried to use the left inverse of $g$ (or right inverse of $f$), but I can't follow it up.

Brian M. Scott
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Gaston Burrull
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  • @JonasMeyer Sorry, I have a mistake when writing question. $g\circ f$ may be an homeomorphism. – Gaston Burrull May 17 '12 at 21:37
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    Do you mean that $f$ and $g$ are functions as above such that $g\circ f$ is a homeomorphism? As it stands it reads that $f$ and $g$ are homeomorphisms. – M.B. May 17 '12 at 21:39
  • @M.B. Yes, I do. A priori $f,g$ only are continuous. But I was edited mistake. – Gaston Burrull May 17 '12 at 21:40
  • OK. Do you see why $g$ is bijective? And how this implies that $f$ is bijective? – M.B. May 17 '12 at 21:42
  • This isn't true. Consider $f$ the identity in $ \mathbb{R}^{n}$ and $g(x) = \exp (x)$. Then, $f,g$ and $g\circ f = g$ are continuous but $g$ isn't homeomorphisms. – user29999 May 17 '12 at 21:47
  • @Marcos: Neither is $g\circ f$, so the example is irrelevant. – Brian M. Scott May 17 '12 at 21:49
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    @Marcos: but the composition is assumed to be a homeomorphism. – M.B. May 17 '12 at 21:49
  • @Marcos composition may be an homeomorphism – Gaston Burrull May 17 '12 at 21:49
  • @M.B. Yes I do with $g$, but I don't see that $f$ is biyective. – Gaston Burrull May 17 '12 at 21:50
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    @AnneBauval: unless the OP can travel into the future, I think she would know been able to see that a similar question appeared 10 years after the original posting. – Mittens Oct 20 '22 at 22:03
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    @Oliver Díaz This was an automatic comment after my vote to close for duplicate. I chose to direct the duplicate in time-reversed direction because the answers to the new post are more complete. – Anne Bauval Oct 20 '22 at 22:09
  • @AnneBauval: Most likely the OP surpassed the level of sophistication required to answer this question several years ago. – Mittens Oct 20 '22 at 22:12
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    @Oliver Díaz I don't care. As already explained, I personally didn't ask any question (the bot did), I am not awaiting any answer, and I just want to link the two posts in the most useful direction. – Anne Bauval Oct 20 '22 at 22:14
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    @AnneBauval: Linking the older the the new one is preferable, unless the newer version is a far more superior question, which is not the case here. – Mittens Oct 20 '22 at 22:18
  • I just got a huge load of notifications for this question I did years ago. – Gaston Burrull Nov 25 '22 at 23:10

2 Answers2

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Suppose that $g$ is injective. It’s also continuous, so it’s a homeomorphism iff it is open. To show that $g$ is open, let $U$ be any open subset of $N$. The map $f$ is continuous, so $f^{-1}[U]$ is open in $M$. The map $g\circ f$ is a homeomorphism, so $g[U]=(g\circ f)[f^{-1}[U]]$ is open in $P$, and therefore $g$ is open.

Note that since $g\circ f$ is injective, $f$ must be injective as well. Thus, to show that $f$ is a homeomorphism, you need only show that it is open. For this you can use the same sort of reasoning as I used above.

Brian M. Scott
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Suppose $g$ is injective.

  • Since $g\circ f$ is onto, $g$ is onto. Since $g$ is a bijection, it has an inverse $g^{-1}$

  • From $Id_P=g\circ f\circ(g\circ f)^{-1}$ it follows that $g^{-1}=g^{-1}\circ g\circ f\circ(g\circ f)^{-1}=f\circ(g\circ f)^{-1}$. Since the right hand side is a composition of continuous functions, $g^{-1}$ is also continuous.

  • Hence, $g$ is a continuous bijection with a continuous inverse, aka a homeomorphism.
  • $f$ is surjective since $f=g^{-1}\circ(g\circ f)$ is a composition of surjections, and injective since $g\circ f$ is injective.
  • As before, $f^{-1}=(g\circ f)^{-1}\circ g$ follows from $Id_M=(g\circ f)^{-1}\circ g\circ f$. Thus, $f^{-1}$ is also continuous.

A similar proof will work if $f$ is assumed to be surjective.

rschwieb
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  • Why $g^{-1}=(f\circ g)^{-1}f$? A priori we don't know if $(fog)^{-1}$ there exist and the next equality holds $(f\circ g)^{-1}=g^{-1}f^{-1}$. – Gaston Burrull May 17 '12 at 21:46
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    Ah! Darn you and your non-alphabetic composition... let me double check to see if it is salvagable. – rschwieb May 17 '12 at 21:49
  • Ok, thanks for effort ;). – Gaston Burrull May 17 '12 at 21:50
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    @GastónBurrull it turns out the answer is easily salvageable, so I'm putting it back. – rschwieb May 17 '12 at 23:29
  • A little detail, you probably must to explain better the equality $g^{-1}=f(gf)^{-1}$ saying that holds because $(gf)^{-1}=(gf)_l^{-1}=f_l^{-1}g_l^{-1}$ where $f_l^{-1}$ and $g_l^{-1}$ are left inverses (and exists) because $g$ and $f$ are 1-1. – Gaston Burrull May 18 '12 at 01:06
  • Remember that $(gf)^{-1}=f_{l}^{-1}g_{l}^{-1}$ or $(gf)^{-1}=f_{r}^{-1}g_{r}^{-1}$ not always holds, we need at least that $g$ and $f$ are at same time or right invertibles or left invertibles. – Gaston Burrull May 18 '12 at 01:21
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    @GastónBurrull It's not necessary to use $f^{-1}$. It follows from $Id_P=g\circ f\circ(g\circ f)^{-1}$ that $g^{-1}=g^{-1}\circ g\circ f\circ(g\circ f)^{-1}=f\circ(g\circ f)^{-1}$. I'll put this in. – rschwieb May 18 '12 at 01:23
  • Thanks for clarification!! – Gaston Burrull May 18 '12 at 02:00