Homeomorphisms and Topologies

• Jan 9th 2013, 06:37 PM
I-Think
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?
• Jan 10th 2013, 03:02 PM
Deveno
Re: Homeomorphisms and Topologies
some technical quibbles:

there is no reason to suppose that for an arbitrary U in X1, that f(X1) ∈ T2.

rather, you want to pick an arbitrary U ∈ T1,and show that f(U) ∈ T2.

(note that saying f(T1) = T2 is actually an abuse of notation, f is a function from X1 to X2, not a function between the respective power sets. there is, however an induced function [f] which does map between the power sets:

[f](U) = {x2 in X2: x2 = f(x1) for some x1 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 X2

(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(T1) is a subset of T2.

on the other hand, suppose V is an open set in X2 (that is, an element of T2). since f is continuous, f-1(V) (here i mean pre-image) is open in X1,

so V = f(f-1(V)) is the image of an open set in X1. this means that T2 is a subset of f(T1).

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