Must a surjective continuous map on a compact space into itself be injective?

Question

Let $X$ be a compact space and $f:X\to X$ a self map on that space. Suppose that $f$ is continuous and surjective. Is it then also injective? Without the compactness condition not necessarily, but is compactness sufficient? If not, what else needs to be demanded?

Answer 1

You can have for instance $X=[0,\pi]$, $f(x)=\pi\sin(x)$.

One condition you can demand is that $X$ is finite. Other than that, unless $X$ is very special there will always be such a function.

Answer 2

$f:[0,1]\rightarrow[0,1]$, $f(x)=-2x+1$ for $x\in[0,1/2]$, $f(x)=2x-1$ for $x\in[1/2,1]$, $f$ is not injective.

Answer 3

Consider $4(x-1/2)^2$ on $X=[0,1].$

Answer 4

A very geometric and natural example from the algebraic topology would be $f:\mathbb{S}^1 \rightarrow \mathbb{S}^1$ s.t. $f(z)=z^2$. Here we define a circle as $\mathbb{S}^1=\{z\in \mathbb{C}=\mathbb{R}^2:|z|=1\}$ Above you see a picture representing this map.

If you got interested, then look for covering maps. They appear quite often, though not often as endomorphisms. One might also ask, could we do the same for higher dimensional spheres $\mathbb{S}^n=\{x\in \mathbb{R}^{n+1}:|x|=1\}$ for $n\geq 2$ ? The answer is no, and in the dimension $1$ we are just lucky that $\mathbb{S}^1=\mathbb{RP}^1$.