Homeomorphisms and Topologies

Wanted to make sure I have all the technical details in the following argument correct

Let $\displaystyle (X_1,T_1)$ and $\displaystyle (X_2,T_2)$ be topological spaces and $\displaystyle f:X_1\rightarrow{X_2}$ be a bijection. Prove that $\displaystyle f$ is a homeomorphism if and only if $\displaystyle f(T_1)=T_2$

Proof

Let $\displaystyle f$ be a homeomorphism. Consider $\displaystyle U\subset{X_1}$ and let $\displaystyle f(U)\in{T_2}$. $\displaystyle f(U)$ is open, and $\displaystyle f$ is continuous, so $\displaystyle U$ is open, so $\displaystyle U\in{T_1}$

And let $\displaystyle U\in{T_1}$, $\displaystyle f(U)\subset{X_2}$, $\displaystyle U$ is open and $\displaystyle f^{-1}$ is continuous, so $\displaystyle f(U)$ is open and $\displaystyle f(U)\in{T_2}$

Now let $\displaystyle f(T_1)=T_2$, so if $\displaystyle U\in{T_1}$, then $\displaystyle f(U)\in{T_2}$, so both are open so $\displaystyle f^{-1}$ is continuous and when $\displaystyle f(U)\in{T_2}$, then $\displaystyle U\in{T_1}$ so $\displaystyle f$ is continuous, thus a homeomorphism.

QED

Are all the details in the proof correct?

Re: Homeomorphisms and Topologies

some technical quibbles:

there is no reason to suppose that for an arbitrary U in X_{1}, that f(X_{1}) ∈ T_{2}.

rather, you want to pick an arbitrary U ∈ T_{1},and show that f(U) ∈ T_{2}.

(note that saying f(T_{1}) = T_{2} is actually an abuse of notation, f is a function from X_{1} to X_{2}, not a function between the respective power sets. there is, however an induced function [f] which does map between the power sets:

[f](U) = {x_{2} in X_{2}: x_{2} = f(x_{1}) for some x_{1} in U} hopefully, no confusion will arise from this, and i will use f(U) to mean [f](U)).

if f is a homeomorphism, then since U is open, and f^{-1} is a continuous function, (f^{-1})^{-1}(U) = f(U) is open in X_{2}

(here the "inner" inverse symbol means "inverse function" and the "outer" inverse symbol means "pre-image", these two sets are equal because f is bijective).

this shows that, at the very least, f(T_{1}) is a subset of T_{2}.

on the other hand, suppose V is an open set in X_{2} (that is, an element of T_{2}). since f is continuous, f^{-1}(V) (here i mean pre-image) is open in X_{1},

so V = f(f^{-1}(V)) is the image of an open set in X_{1}. this means that T_{2} is a subset of f(T_{1}).

the other implication you prove looks ok, i would just note that both f and f^{-1} are open maps, so both are continuous.