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

[docnet]

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.

 

[PeroK]

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.

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!

 

[PeroK]

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?

 

[WWGD]

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

 

[pasmith]

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.

 

[PeroK]

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!