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- 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:

##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.

##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.

**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.