Question

Prove or disprove : A continuous bijection is a homeomorphism

Answer 1

$\text{Let I=[a,b] be a closed interval with a<b then without loss of generality I can assume}\\\text{f(a)<f(b), I want to prove that f|I is strictly increasing. If not, there exists x<y with}\\\text{f(x)>f(y) by injectivity can't be the equality. If f(a)<f(y)<f(x) intermediate value}\\\text{theorem give me c∈(a,x) such that f(c)=f(y) but it can't be by injectivity. If f(y)<f(a)}\\\text{is the same. Since a continuous} \\\text{injective function $f:\mathbb{R}\to\mathbb{R} $ is strictly increasing or } \\\text{decreasing follows from the fact that if f is strictly increasing in any closed interval,}\\\text{ obviously f $\in \mathbb{R} $ Then f is strictly increasing. If image is all, I'll prove} \\\text{that $f^{-1}=g$ is continuous. It is well known that if f is strictly increasing then so is g.}\\\text{I'll prove continuity in some a∈R. Let ϵ>0 then g(a)−ϵ/2 and g(a)+ϵ/2 have}\\\text{preimages let g(b)=g(a)−ϵ/2 and g(c)=g(a)+ϵ/2 then b<a<c, Since a continuous}\\\text{injective function $f:\mathbb{R}\to\mathbb{R} $ is strictly increasing or decreasing then we obtain}\\\text{δ=min{|b−a|,|c−a|} then g is continuous, therefore f is a homeomorphism.}$