Exercise 6.2.21:

  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.

  • 1
    $\begingroup$ I suppose you mean $H_A (B) \cong X (B) + Y (B)$. Take $B = A$ and see what happens to $\mathrm{id}_A$. $\endgroup$ – 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.


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