# Prove there is no bijective continuous function $f:[0,1] \to [0,1]^2$ [duplicate]

Prove there is no bijective continuous function $f:[0,1] \to [0,1]^2$.

Here is my attempt so far: assume there a function $[0,1] \to [0,1]^2$ is continuous and surjective. We wish to show that it cannot be injective. Suppose in the contrary that it is injective. Then since the preimage of $f$ is compact and $f$ is bijective, its inverse $f^{-1}$ is a bijection from the unit square to $[0,1]$. Also, it is a uniform convergence.

I am also wondering if there is a not necessarily continuous bijection from the unit square to unit line.

I am not sure how to prove this problem. Any help would be great.

## marked as duplicate by user99914, Namaste, Foobaz John, Community♦Feb 11 '18 at 0:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

A bijective continuous function would be a closed map (maps closed sets to closed sets) and hence a homeomorphism. To see the map is closed note that if $C$ is closed in $[0,1]$ it is compact, then $f(C)$ is compact as $f$ is continuous whence $f(C)$ is closed as compact sets in a Hausdorff space are closed.

But $[0,1]$ and $[0,1]^2$ are not homeomorphic. Remove a point from $[0,1]$ and the space is disconnected while removing a point from $[0,1]^2$ leaves it connected.