# Showing that a functor is not representable.

Let $Y,Z$ be sets. We define a functor $F:Set\rightarrow Set$ in the following manner:

$F(X)=Hom(X,Y)\amalg Hom(X,Z)$, $F(f)=Hom(X,f)\amalg Hom(X,f)$.

How to prove that $F$ is not representable unless one of $Y,Z$ is empty?

Thank you in anticipation.

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Probably no-one is responding because it is easy for you all, but a solution will be greatly appreciated, thank you for your time. – keka Dec 4 '13 at 7:43
Does $\amalg$ mean disjoint union? And what is $X$ in $F(f) = Hom(X, f) \amalg Hom(X, f)$. Should this be $Hom(Y, f) \amalg Hom(Z, f)$? – Aleš Bizjak Dec 4 '13 at 8:00
Or perhaps even $\mathrm{Hom}(f, Y) \amalg \mathrm{Hom}(f, Z)$? – Zhen Lin Dec 4 '13 at 8:16

If $F = \hom(-,Y) \sqcup \hom(-,Z) \cong \hom(-,S)$ for some set $S$, then by plugging in $1$ (the set with one element) we get $Y \sqcup Z = S$. It follows that the natural map $\hom(-,Y) \sqcup \hom(-,Z) \to \hom(-,Y \sqcup Z)$ is an isomorphism. In other words, if $X \to Y \sqcup Z$ is an arbitrary map, then its image lies in $Y$ or in $Z$. Of course this fails when $Y,Z \neq \emptyset$ (consider $X=2$).