Does the Yoneda embedding tell us the set of all maps from a compact genus zero surface to the sphere is isomorphic to SO(3)? I am just getting my head around the Yoneda embedding (and lemma) and want to check if my understanding is correct.
The surface classification theorem says that every closed orientable compact genus zero surface $X$ is homeomorphic to $S^2$. If we consider the Yoneda embedding of the category of closed surfaces where morphisms are homeomorphisms, then $S^2 \mapsto Hom(S^2,S^2)$ and $X \mapsto Hom(S^2,X)$. But $Hom(S^2,S^2) \cong SO(3)$ so does it mean that $Hom(S^2,X) \cong SO(3)$?

In any (locally small) category $\mathscr C$, if there is an isomorphism between $a$ and $b$, then the presheaves $\hom_{\mathscr C}(-,a)$ and $\hom_{\mathscr C}(-,b)$ are isomorphic. You don't need Yoneda's lemma for that : it just follows from the fact that $c \mapsto \hom_{\mathscr C}(-,c)$ is a functor.
So, working in the category of surfaces and continuous maps where $X$ and $S^2$ are isomorphic, you end up with a isomorphism between $\hom(-,S^2)$ and $\hom(-,X)$. Of course, it gives you a isomorphisms of sets (i.e. a bijection)
$$ \hom(S^2,S^2) \simeq \hom(S^2,X). $$
So it just says that $\hom(S^2,X)$ has the same cardinality as $\hom(S^2,S^2)$. If you have a group structure you like on $\hom(S^2,S^2)$, feel free to transport it to $\hom(S^2,X)$, but this isn't news (I mean you can put that structure on any other set of the right cardinality).
