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From Steve Awodey "Category Theory" Second Edition (2010), page 63:

where the last line really means that there is a factorization $z = \bar{z} ◦i$ of z through the inclusion $i : S → R^2$, as indicated in the following diagram commutative diagram

It seems that the proof that $\bar{z}$ is a $\mathbf{Top}$-morphism (that is a continuous function) is missing.

How to show that here $\bar{z}$ is a continuous function?

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Awodey says on the previous page that he's working in $\mathbf{Top}$. I think the question to ask is how he can be sure that equalizers necessarily exist in $\mathbf{Top}$ and that $S$ is in fact an equalizer of $\langle x,y\rangle\mapsto x^2+y^2$ and $\langle x,y\rangle\mapsto 1$. – MJD Oct 14 '13 at 21:01
up vote 5 down vote accepted

Actually the proof is provided some lines before indeed Awodey says:

For, given any generalized element $z \colon Z \to \mathbb R^2\dots$ we get a pair of such elements $z_1,z_2 \colon Z \to \mathbb R$ just by composing with the two projections, $z= \langle z_1,z_2\rangle$, for these we have that $f(z)=g(z)$ iff $z_1^2+z_2^2=1$ iff $z \in S$.

What this actually tells is that the generalized element (i.e. the continuous function) $z \colon Z \to \mathbb R^2$ to equalize the two morphisms $f$ and $g$ must have image contained in $S$ and so it have to factors through a continuous function and the embedding $i \colon S \to \mathbb R^2$.

What's actually used is a general fact about topological spaces:

For every map in $\mathbf {Top}$, let's say $f \colon X \to Y$, if such map have image contained in $A \subseteq Y$ then this maps factors uniquely through the inclusion $i \colon A \to Y$, i.e. there's a continuous function $\bar f \colon X \to A$ such that $f= i \circ \bar f$.

This statement is a topological one, and it's one of the properties of topological subspace. It's clear that the factorization exists in $\mathbf{Set}$, the fact that the map is continuous come from the fact that $A$ is not just a subset of $X$ but it's a subspace, meaning that it's a topological space with topology induced from $Y$ (the open sets of $A$ are the intersections of open sets of $Y$ with $A$). From this it follows that the for every open set $U$ of $A$ the set $\bar f^{-1}(U)$ must be equal to some $f^{-1}(U')$ for some $U'$ open set of $Y$ such that $U' \cap A=U$.

If I'm right this is also a category-theoretic (i.e. element free) characterization of topological subspaces.

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Please be clear: your second blockquote, is it from Awodey or is it your own passage? – porton Oct 14 '13 at 21:35
@porton no it's not in Awodey's book, it's mine :) – Giorgio Mossa Oct 14 '13 at 21:36
@porton My apologize, I've added some details I hope now I've made myself more clear. If that's not the case feel free to ask.:) – Giorgio Mossa Oct 15 '13 at 10:39
"follows that the for every open set" - the word "the" is superfluous – porton Oct 15 '13 at 11:41
I'm not quite sure that I understand why $\bar f^{-1}(U) = f^{-1}(U')$ – porton Oct 15 '13 at 11:54

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