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In Lee's book he uses the characteristic property of the subspace topology:

Suppose X is a topological space and $S \subset X$ is a subspace. For any topological space Y, a map $f: Y \rightarrow S$ is continuous iff $r_S \circ f : Y \rightarrow X$ is continuous, where $r_S$ is the inclusion map $r_S : S \rightarrow X$

Now to show that the inclusion map $r_S : S \rightarrow X$ is continuous he argues by saying that the identity is always continuous and uses the fact that the following diagram commutes:

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Which to me looks as if he's proving $r_S$ is continuous by arguing that $r_S$ is continuous, which is probably not what he means.

So I think that I must not completely understand the usefulness of the characteristic property theorem. Could someone help me maybe by explaining his proof please?

eager2learn
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    No, he is arguing that $r_s$ is continuous because $id$ is. Put every player into the universal property (i.e. $Y=S$ and $f=id$) and see that its all ok. – freakish Oct 08 '18 at 15:36

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Apply the universal property for $Y= S$ and $f=\mathrm{Id}_S$: it says

$\mathrm{Id}_S$ is continuous iff $r_S \circ \mathrm{Id}_S$ is continuous.

The latter is just $r_S$ by the commutative diagram, which is trivial to check.

And the identity on space is always continuous (just as all constant maps), so we conclude that as the left function is continuous, so is $r_S$.

Henno Brandsma
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