In Tom Leinster's Basic Category Theory, it says that the coproduct of two representable functors is never representable, but I cannot see why this is true. Suppose we have two Hom-functors $h_a$ and $h_b$ and the coproduct $a+b$ exists. Then, $h_a+h_b$ is to be computed pointwise, i.e. for any object $c$, $(h_a+h_b)(c) = h_a(c)+h_b(c)$, which is just the disjoint union of the Hom-sets from $c$. Now how is this different from $h_{a+b}$? The injections of $a$ and $b$ into the coproduct make $\text{Hom}(c,a)+\text{Hom}(c,b)$ isomorphic to $\text{Hom}(c,a+b)$ as sets, don't they? What am I overlooking here?

Please provide an argument that shows why the coproduct of two representables is never representable.

  • 3
    $\begingroup$ No, it's $Hom(a,c)\times Hom(b,c)$ that is isomorphic to $Hom(a+b,c)$. Your isomorphism doesn't exist in general (think in terms of finite sets : $hom(c,a)$ is $a^c$ and $hom(c,b)$ is $b^c$: you're saying $a^c + b^c = (a+b)^c$ $\endgroup$ Aug 10, 2018 at 14:07
  • $\begingroup$ Have you tried seeing if what you say is true for the category of sets, for example? $\endgroup$
    – Arnaud D.
    Aug 10, 2018 at 14:08
  • $\begingroup$ Alright, so now I see why this is not true in general. However, can someone please provide a proof why the coproduct of two representables is never representable? $\endgroup$
    – wave
    Aug 10, 2018 at 14:30
  • 4
    $\begingroup$ An (overly) sophisticated way for seeing that this is likely to be true is to note that the category of presheaves, i.e. $[\mathcal C^{op},\mathbf{Set}]$ (for small $\mathcal C$) is (with the Yoneda embedding) the free cocompletion of $\mathcal C$. This means we freely attach a colimit for every diagram (identifying them when they should be the "same"). This suggests that you generally shouldn't expect the colimit of a non-degenerate diagram to give rise to a presheaf in the image of $\mathcal C$. For (non-trivial) coproducts, there is nothing that could possibly force them to coincide. $\endgroup$ Aug 14, 2018 at 5:43

1 Answer 1


Assume that for some objects $a,b,c$ of $\mathcal{C}$ we have $h_a+h_b\cong h_c$. Then for any functor $F:\mathcal{C}^{op}\to \mathbf{Set}$, we have by definition of the coproduct $h_a+h_b$ $$\operatorname{Nat}(h_a,F)\times \operatorname{Nat}(h_b,F)\cong \operatorname{Nat}(h_a+h_b,F)\cong \operatorname{Nat}(h_c,F),$$ and then by the Yoneda lemma $$F(a)\times F(b)\cong F(c).$$ But clearly this cannot be true for the functor $F$ that maps every object to the constant set $\{0,1\}$ (or any finite set with more than one element) and every arrow to the identity, so we have a contradiction.

  • 1
    $\begingroup$ In the question the representables are contravariant, though. $\endgroup$
    – asdq
    Aug 10, 2018 at 14:51

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