Thread: Homeomorphisms and Topologies

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

Let and be topological spaces and be a bijection. Prove that is a homeomorphism if and only if

Proof
Let be a homeomorphism. Consider and let . is open, and is continuous, so is open, so
And let , , is open and is continuous, so is open and

Now let , so if , then , so both are open so is continuous and when , then so is continuous, thus a homeomorphism.
QED

Are all the details in the proof correct?

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.