The Stacks project

Proof. If $F$ is representable, then it commutes with colimits by definition of colimits.

Assume that $F$ commutes with colimits. Then $F(M \oplus N) = F(M) \times F(N)$ and we can use this to define a group structure on $F(M)$. Hence we get $F : \mathcal{A} \to \textit{Ab}$ which is additive and right exact, i.e., transforms a short exact sequence $0 \to K \to L \to M \to 0$ into an exact sequence $F(K) \leftarrow F(L) \leftarrow F(M) \leftarrow 0$ (compare with Homology, Section 12.7).

Let $U$ be a generator for $\mathcal{A}$. Set $A = \bigoplus _{s \in F(U)} U$. Let $s_{univ} = (s)_{s \in F(U)} \in F(A) = \prod _{s \in F(U)} F(U)$. Let $A' \subset A$ be the largest subobject such that $s_{univ}$ restricts to zero on $A'$. This exists because $\mathcal{A}$ is a Grothendieck category and because $F$ commutes with colimits. Because $F$ commutes with colimits there exists a unique element $\overline{s}_{univ} \in F(A/A')$ which maps to $s_{univ}$ in $F(A)$. We claim that $A/A'$ represents $F$, in other words, the Yoneda map

is an isomorphism. Let $M \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $s \in F(M)$. Consider the surjection

This gives $F(c_ M)(s) = (s_\varphi ) \in \prod _\varphi F(U)$. Consider the map

which maps the summand corresponding to $\varphi $ to the summand corresponding to $s_\varphi $ by the identity map on $U$. Then $s_{univ}$ maps to $(s_\varphi )_\varphi $ by construction. in other words the right square in the diagram

commutes. Let $K = \mathop{\mathrm{Ker}}(A_ M \to M)$. Since $s$ restricts to zero on $K$ we see that $\psi (K) \subset A'$ by definition of $A'$. Hence there is an induced morphism $M \to A/A'$. This construction gives an inverse to the map $h_{A/A'}(M) \to F(M)$ (details omitted). $\square$