# If $g∘f$ is a homemorphism, then $g$ one-one (or $f$ onto) implies that $f$ and $g$ are homeomorphisms.

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?

• 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. Commented Oct 17, 2022 at 23:40
• Yes, you're right.
– Paul
Commented Oct 17, 2022 at 23:41

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.
• Thank you, it helps me a lot. I understand everything at points 1 - 5. Commented Oct 19, 2022 at 17:17

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.
• Since this answer has just been downvoted, how could I improve it to regain your esteem? Commented Oct 18, 2022 at 9:48
• Again today: why? Commented Oct 27, 2022 at 14:35