Suppose $\mathcal{C}$ is a locally small category with coproducts, and there is a functor $G: \mathcal{C} \to \operatorname{Set}$ which is representable, with representation $(A, x)$. I am trying to show that $G$ has a left adjoint, and I have been given a clue that this is the functor $F : \operatorname{Set} \to \mathcal{C}$ where on objects, $F(I) = \displaystyle\coprod_{i\in I} A$, the coproduct of $|I|$ copies of $A$. However, I am not sure where a function $f:I \to J$ is mapped to by this functor. Can anyone give me an idea? Thanks for any help.

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    $\begingroup$ Define $F(f)$ by mapping the $i$th copy of $A$ in $\amalg_i A$ to the $f(i)$th copy of $A$ in $\amalg_j A$. $\endgroup$ – Arturo Magidin May 10 '12 at 17:05
  • $\begingroup$ So I think this means the unique map $F(f) : \coprod_i A \to \coprod_j A$ such that $F(f) \circ \iota_i = \iota_{f(i)}$, where the $\iota$s are the maps into the coproduct? $\endgroup$ – Paul Slevin May 10 '12 at 17:15
  • $\begingroup$ Correct. Although if you want to be finicky about notation you may as well use different symbols for the different coproducts. $\endgroup$ – Zhen Lin May 10 '12 at 17:16
  • $\begingroup$ of course, thanks $\endgroup$ – Paul Slevin May 10 '12 at 17:23
  • $\begingroup$ @PaulSlevin Maybe you want to answer your own question with the hints in the comments, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 15 '13 at 12:53

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