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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?

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    $\begingroup$ 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 $\mathbb{S}^1=\{z\in \mathbb{C} = \mathbb{R}^2 : |z|=1 \}$. Here is the picture $\endgroup$ – Mihail Mar 11 '18 at 11:20
  • $\begingroup$ @Mihail Your comment deserves to be an answer. And I'd prefer to upvote it to upvoting the comment $\ddot\smile$ $\endgroup$ – Hanno Mar 11 '18 at 13:02
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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.

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

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  • $\begingroup$ thanks, the tent map, or upside down tent map, is of course a good example. $\endgroup$ – Marlo Mar 11 '18 at 0:20
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Consider $4(x-1/2)^2$ on $X=[0,1].$

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    $\begingroup$ As someone who knows much less about topology than the asker: would this be the only counterexample, one of many counterexamples of the only available type, or just one counterexample of many different kinds of counterexamples? $\endgroup$ – Robert Soupe Mar 10 '18 at 19:38
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    $\begingroup$ @RobertSoupe It's just one example of zillions $\endgroup$ – zhw. Mar 10 '18 at 19:59
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enter image description here

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 $\mathbb{S}^1=\mathbb{RP}^1$.

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  • $\begingroup$ The linked pic is - only currently? - not available. Why not including it into your answer? $\endgroup$ – Hanno Mar 11 '18 at 20:59
  • $\begingroup$ @Hanno Thank you for both comments $\endgroup$ – Mihail Mar 11 '18 at 22:29

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