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I need a little explanation, please. Seymour Lipschutz - General Topology, Chapter 7, page 110:

Let $X,Y,Z$ be topological spaces and let $f:X\longrightarrow Y$ and $g:Y\longrightarrow Z$ be continuous. Show that if $g\circ f:X\longrightarrow Z$ is a homemorphism, then $g$ one-one (or $f$ onto) implies that $f$ and $g$ are homeomorphisms.

To show that two functions $f$ and $g$ are homeomorphisms we need to show that there exists an one-to-one correspondence, and $f,f^{-1},g,g^{-1}$ are continuous. As $f$ and $g$ are already continuous by hypothesis, and $g$ one-one (or $f$ onto), we only have to show that $g$ onto (or $f$ one-one), $f^{-1}$, and $g^{-1}$ are continuous. Am I right?

Anne Bauval
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Tryncha
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  • Maybe you should contemplate the example $X=Z={0}$ and $f,g$ both constant functions taking the value $0$. Then neither $f$ nor $g$ are bijective, but $g\circ f$ is a homeomorphism. To be less oblique-you need to show that $f$ and $g$ are invertible, as well as that they have continuous inverses. – Jamie Radcliffe Oct 17 '22 at 23:40
  • Yes, you're right. – Paul Oct 17 '22 at 23:41

2 Answers2

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As $f$ and $g$ are already continuous by hypothesis, and $g$ one-one (or $f$ onto), we only have to show that $g$ onto (or $f$ one-one), $f^{-1}$, and $g^{-1}$ are continuous. Am I right?

Yes, you're right. Alternatively, you can show that $f$ and $g$ are bijective and open (since they are continuous, it will imply they are homeomorphism).

I consider the case where $g$ is one-to-one, the case of $f$ being similar. There are a few steps, I let you check the details:

  1. $g\circ f$ is a homeomorphism, so it is one-to-one. This implies that $f$ is one-to-one
  2. Similarly, $g\circ f$ is onto, so $g$ is onto.
  3. $g$ is also one-to-one by assumption, so $g$ is bijective.
  4. $g\circ f$ and $g$ are bijective, so $f=g^{-1}\circ(g\circ f)$ is also bijective.
  5. Since $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$, we have $f^{-1}=(g\circ f)^{-1}\circ g$.

Note that points $1-5$ are standard facts from Set Theory. If you don't know them, I recommend you to check them, as they are useful for topology.

  1. Finally, let's show that $f$ is open. Let $O$ be open in the topology $\mathcal O_X$ of $X$. Then, since $g$ is continuous and $g\circ f$ is open,

$$ f(O) = g^{-1}\left( g\circ f(O) \right) $$

is also open. Therefore, $f$ is a homeomorphism.

  1. I follows immediately that $g=(g\circ f)\circ f^{-1}$ is a homeomorphism.
Taladris
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Yes, you are right. Since the first case (where $g$ is one-to-one) was already solved in this duplicate, let us solve the second case, though it is very similar.

So, suppose $f$ is onto.

  • Since $g\circ f$ is one-to-one, so is $f.$ Since $f$ is a bijection, it has an inverse $f^{-1}.$
  • From ${\rm id}_Z=(g\circ f)^{-1}\circ g\circ f$ it follows that $f^{-1}=(g\circ f)^{-1}\circ g\circ f\circ f^{-1}=(g\circ f)^{-1}\circ g.$ Since the right hand side is a composition of continuous functions, $f^{-1}$ is also continuous.
  • Hence, $f$ is a continuous bijection with a continuous inverse, i.e. a homeomorphism.
  • Finally, $g=(g\circ f)\circ f^{-1}$ is a composition of two homeomorphisms, hence also a homeomorphism.
Anne Bauval
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