# Prove that f is a homeomorphism iff g is continuous, fg=1 and gf=1

• docnet
In summary, the proof shows that if a function is a homeomorphism, then it has a continuous inverse. However, the converse is not necessarily true, as there may exist continuous functions with continuous inverses that are not homeomorphisms.
docnet
Gold Member
Homework Statement
Prove that ##f:X\longrightarrow Y
## is a homeomorphism if, and only if, there exists a continuous map ##g:Y\longrightarrow X## so that ##fg=1## and ##gf=1## are both the identity.
Relevant Equations
none.
Outline of proof:

Part I:

##1.## ##f## is a homeomorphism, so there exists a continuous inverse ##g:Y\longrightarrow X##.

##2.## ##f## is a bijection, hence there is a unique ##f(x)## in ##Y## for every ##x## in ##X##. For every ##f(x)\in Y##, the preimage under ##f## is ##f^{-1}f(x)=x=gf(x)##, so ##gf=1## is the identity.

##3.## There is a unique ##g(y)## in ##X## for every ##y## in ##Y##. For every ##g(y)## in ##X##, the preimage under ##g## is ##g^{-1}g(y)=y=fg(y)##, so ##fg=1## is the identity.

Thus, we have found a continuous ##g:Y\longrightarrow X## such that ##fg## and ##gf## are the identity.

Part II:

##4.## We are given ##g## is continuous and ##f## is arbitrary, so that ##fg=1## and ##gf=1## are both the identity .

##5.## It follows from earlier results (that follows from ##4.##) that ##g## is the continuous inverse of ##f## and it is bijective.

##6.## Continuous functions form a ##\mathbb{R}##-algebra, so ##f## is a continuous function, hence ##f## is a homeomorphism.

docnet said:
Homework Statement:: Prove that ##f:X\longrightarrow Y
## is a homeomorphism if, and only if, there exists a continuous map ##g:Y\longrightarrow X## so that ##fg=1## and ##gf=1## are both the identity.
Relevant Equations:: none.

Outline of proof:

Part I:

##1.## ##f## is a homeomorphism, so there exists a continuous inverse ##g:Y\longrightarrow X##.

##2.## ##f## is a bijection, hence there is a unique ##f(x)## in ##Y## for every ##x## in ##X##. For every ##f(x)\in Y##, the preimage under ##f## is ##f^{-1}f(x)=x=gf(x)##, so ##gf=1## is the identity.

##3.## There is a unique ##g(y)## in ##X## for every ##y## in ##Y##. For every ##g(y)## in ##X##, the preimage under ##g## is ##g^{-1}g(y)=y=fg(y)##, so ##fg=1## is the identity.

Thus, we have found a continuous ##g:Y\longrightarrow X## such that ##fg## and ##gf## are the identity.
Part I is true by definition. There's nothing to prove.
docnet said:
Part II:

##4.## We are given ##g## is continuous and ##f## is arbitrary, so that ##fg=1## and ##gf=1## are both the identity .

##5.## It follows from earlier results (that follows from ##4.##) that ##g## is the continuous inverse of ##f## and it is bijective.

##6.## Continuous functions form a ##\mathbb{R}##-algebra, so ##f## is a continuous function, hence ##f## is a homeomorphism.
I'm not convinced by that. Clearly ##g## is an inverse for ##f##, so ##f## is a bijection (one-to-one and onto) - I don't think that needs a proof.

The thing that needs proving is that ##f## must be continuous. That is the one thing that is not obvious about the proposition.

I suggest you need to do more than quote something about an ##\mathbb{R}##-algebra!

docnet said:
Homework Statement:: Prove that ##f:X\longrightarrow Y
## is a homeomorphism if, and only if, there exists a continuous map ##g:Y\longrightarrow X## so that ##fg=1## and ##gf=1## are both the identity.
Where did you get this question?

jim mcnamara
You may want to find an exception to the claim that every continuous injection has a continuous inverse.

docnet said:
Homework Statement:: Prove that ##f:X\longrightarrow Y
## is a homeomorphism if, and only if, there exists a continuous map ##g:Y\longrightarrow X## so that ##fg=1## and ##gf=1## are both the identity.

You being asked to show that if $f$ is a homeomorphism then its inverse is continuous. But isn't a homeomorphism by definition a continuous map with a continuous inverse? If not, what definition of homeomorphism are you using? You don't state it in your post.

pasmith said:
You being asked to show that if $f$ is a homeomorphism then its inverse is continuous. But isn't a homeomorphism by definition a continuous map with a continuous inverse? If not, what definition of homeomorphism are you using? You don't state it in your post.
That implication follows from the definition, but not the converse. Which, despite being proved by the OP, appears not to hold!

## 1. What is a homeomorphism?

A homeomorphism is a type of function that preserves the topological structure of a space. This means that it maps points in one space to points in another space in a way that preserves the relationships between those points.

## 2. How do you prove that f is a homeomorphism?

To prove that f is a homeomorphism, you must show that it is bijective, continuous, and has a continuous inverse function. This means that for every point in the domain of f, there must be a unique point in the range of f, f must be continuous (meaning that small changes in the input result in small changes in the output), and the inverse function of f must also be continuous.

## 3. What does it mean for fg=1 and gf=1?

When fg=1 and gf=1, it means that f and g are inverse functions of each other. This means that when f is applied to a point, and then g is applied to the result, the original point is obtained. Similarly, when g is applied to a point, and then f is applied to the result, the original point is obtained. This is necessary for a function to be a homeomorphism.

## 4. Why is it important for g to be continuous?

G being continuous is important because it ensures that the inverse function of f is also continuous, which is necessary for a function to be a homeomorphism. Additionally, continuity of g is important for ensuring that the composition of f and g is well-defined and meaningful.

## 5. Can you give an example of a homeomorphism?

One example of a homeomorphism is the function f(x) = 2x, which maps the real line to itself. This function is bijective, continuous, and has a continuous inverse (f^-1(x) = x/2), making it a homeomorphism. It preserves the topological structure of the real line, as small changes in the input result in small changes in the output.

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