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I was trying to prove that there exists a continuous function from a space to many others, and to see in what cases there exists such a function.

Clearly, the first things to note are properties preserved by continuous functions, like compactness and connectivity. Are there more general criteria?

I tried to prove or disprove the existence of a continuous function from the sphere (the boundary) to the circle (also the boundary).

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It's easy to see that there is a surjective continuous map from $[0,1]$ to $\mathbb S^1$. Can you see how to make a continuous surjection from $\mathbb S^2$ to $[0,1]$? – JSchlather Aug 30 '11 at 18:00
the construction of such function ( S^2 to [0,1] ) it´s geometrically? or it´s only a non special function? Dx? – Daniel Aug 30 '11 at 18:54
@Daniel: e.g. $(x,y,z) \mapsto |z|$. Quite geometrical I'd say :) – t.b. Aug 30 '11 at 19:03
If you want the most general answer, it will have to be "A surjective continuous mapping exists from $X$ onto $Y$ if and only if there is a function from $X$ which is onto $Y$ and the preimage of every open set in $Y$ is open in $X$". Of course if you have some limitations to pose the answer can be less tautological and with more content in it. – Asaf Karagila Aug 30 '11 at 22:20

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That depends a great deal on the kind of spaces you're interested in. For manifolds, some of the main obstructions are invariants from algebraic topology (homology and homotopy groups) which in a sense generalize the "connectivity" property you already observed is useful.

For more general topological or even metric spaces, the tools (and the answers) are quite different. For example, every compact metric space is a quotient of the Cantor set; in other words you can map the Cantor set surjectively onto pretty much any (compact) thing, just because it's so disconnected it imposes virtually no restrictions on maps from it.

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In fact all the separable metric spaces are continuous image of a closed subset of the Baire space. – Asaf Karagila Aug 30 '11 at 22:41

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