2
$\begingroup$

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.

  • 0
    @M.B. Yes I do with $g$, but I don't see that $f$ is biyective.2012-05-17

2 Answers 2

5

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.

  • 0
    @Gastón: Thank you; I try hard to be clear.2012-05-17
4

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.

  • 0
    Thanks for clarification!!2012-05-18