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I have this problem which seems trivial, but I do not know how to get to the answer. I have the set $[0,1]$ with the Euclidean topology, and an arbitrary continuous function $f : I \rightarrow I^2 = I \times I$. I need to prove that if it is surjective then it cannot be injective.

I tried to think to the problem in this way: if $f$ is surjective and also injective, than we would have a homeomorphism between $I$ and $I \times I$.

I wanted to say that one is compact and the other is not to reach a contradiction. However in my opinion they are both compact because product of compacts is compact as well, so I do not see why it would be a problem to have the injective property. Any tips?

Theoretical Economist
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qcc101
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  • I don't think there is a reasonable notion of 'generic' for which a generic function is surjective. In fact, I believe (but I'm not completely sure) that a generic function will not be surjective (in the sense that there is an open dense set of such functions). You probably meant to write 'arbitrary'. – tomasz Jan 07 '18 at 11:19
  • @tomasz Yep, edited! – qcc101 Jan 07 '18 at 11:23

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Hint: rather than compactness, you might want to look at connectedness. Now both $I$ and $I \times I$ are connected, so that doesn't do much. But what happens if you remove a point from each of them?

John Hughes
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As I is compact Hausdorf, f is closed map.
Thus were f a bijection, it would be a homeomorphism.
As I and I×I are not homeomorphic, f cannot be bijective.
They are not homeomorphic because I has cut points and IxI does not.
A cut point is a point whose removal will disconnect the space.

William Elliot
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