# The connectedness property of representables

1. Prove that representables have the following connectedness property: given a locally small category $$\mathscr A$$ and $$A\in\mathscr A$$, if $$X,Y\in[\mathscr A^{op},\mathbf {Set}]$$ with $$H_A\simeq X+Y$$, then either $$X$$ or $$Y$$ is the constant functor.
2. Deduce that the sum of two representables is never representable.
1. Any hints about this? I've been trying to apply various results haphazardly, but I don't see a plan of the proof that I should follow. What I've tried: since colimits in functor category can be computed pointwise, we have $$H_A(B)\simeq X(B)\times Y(B)$$ naturally in $$B$$. Writing out the condition on the commutativity of the square didn't give anything. Also the presence of $$X(B)$$ alludes to the Yoneda lemma: $$X(B)\simeq [\mathscr A^{op},\textbf {Set}](H_B, X)$$, but I don't see how to apply this either.

2. Suppose $$X$$ and $$Y$$ are representable and suppose their sum is representable. Then by 1, $$X$$ or $$Y$$ is the constant functor. Does this contradict the fact that $$X$$ and $$Y$$ are representable? (I.e., does there no exist constant representable functors?) I don't think so, $$\mathscr A$$ may be a discrete category with 1 element, and then $$H_A$$ will be a constant functor. But then I don't know how to find a contradiction to 1.

• I suppose you mean $H_A (B) \cong X (B) + Y (B)$. Take $B = A$ and see what happens to $\mathrm{id}_A$. – Zhen Lin Jul 4 '20 at 22:34

You might use that a representation of the presheaf $$X+Y$$ corresponds, through the natural bijection of Yoneda Lemma, to a universal element of $$X+Y$$. I mean that, if $$a\in (X+Y)A=XA \sqcup YA$$ is the image of the isomorphism $$H_A \cong X + Y$$ through Yoneda Lemma's bijection: $$Set^{\mathcal{A}^{op}}(H_A,X + Y)\cong (X+Y)A,$$ then the couple $$(A,a)$$ is initial in the category of elements of $$X+Y$$. This means that, whenever $$B$$ is an object of $$\mathcal{A}$$ and $$b \in (X+Y)B$$, then there is unique an arrow $$B \xrightarrow{f}A$$ of $$\mathcal{A}$$ such that the function $$(X+Y)f=Xf \sqcup Yf$$ sends $$a$$ to $$b$$ (this is Corollary 4.3.2 of your link).
If -without loss of generality- we assume that $$a \in XA$$ then, if such a $$b \in (X+ Y)B=XB\sqcup YB$$ exists, it needs to belong to $$XB$$, for the map $$Xf \sqcup Yf$$ sends elements of $$XA$$ to elements of $$XB$$ and elements of $$YA$$ to elements of $$YB$$. This implies that $$YB$$ is empty (and $$B$$ was arbitrary), hence $$Y$$ constantly equals the empty set. If we assumed that $$a \in YA$$ then it would be the case that $$X$$ constantly equals the empty set.
Now 2. is easier, knowing that one between $$X$$ and $$Y$$ is not just a constant presheaf but also the constantly empty one.