Let $\mathcal Z$ be a small category and $\mathcal C$ any category. We then consider the category $\text{Fun}(\mathcal Z, \mathcal C)$ the category of functors from $\mathcal Z$ to $\mathcal C$ and the functor $\Gamma: \mathcal C \to \text{Fun}(\mathcal Z, \mathcal C)$ such that
$$
\Gamma(C):
\begin{pmatrix}
\mathcal Z &\longrightarrow &\mathcal C\\
Z & \longmapsto & C\\
f & \longmapsto & 1_C
\end{pmatrix}
$$
where $f: Z \to Z'$ and $1_C$ is the identity morphism on $C$.
I am trying to show that $\Gamma$ has a left adjoint if and only if the colimit of every functor $F: \mathcal Z \to \mathcal C$. The "if" part is not too hard but I have some difficulties to show the "only if" part.
Here is my idea: Suppose that there is $\Omega : \text{Fun}(\mathcal Z, \mathcal C) \to \mathcal C$ such that
$$\theta_{F, C}: \text{Hom}_\mathcal C(\Omega(F), C)\cong \text{Nat}(F, \Gamma(C)).$$
For each $\alpha \in \text{Nat}(F, \Gamma(C))$ we can associate a cocone, i.e. for $f: Z \to Z'$ the following diagram commutes
$$
\begin{array}{ccc} F(Z)&\xrightarrow{F(f)} &F(Z')\\ \searrow&&\swarrow\\
&\Gamma(C)(Z) = C = \Gamma(C)(Z')& \end{array}
$$
where the arrows $F(Z) \to C$ and $F(Z') \to C$ are given by $\alpha_Z$ and $\alpha_{Z'}$ respectively. Because of the above isomorphism, we can associate to $\alpha$ a unique morphism $\theta^{-1}_{F, C}(\alpha) =u: \Omega(F) \to C$ so $\Omega(F)$ is a good candidate to be the colimit of $F$. We just have to find a family of morphism $\mu_Z: F(Z) \to \Omega(F)$ such that $\alpha_Z = u \circ \mu_Z$. My guess is that this family of morphism is given by the unit of the adjunction $\eta_F \in \text{Nat}(F, \Gamma(\Omega(F)))$ but I am not able to show that
$$\alpha_Z = u \circ (\eta_F)_Z = \theta^{-1}_{F, C}(\alpha) \circ (\theta_{F, \Omega(F)}(1_{\Omega(F)}))_Z$$
where the last equality comes from the expression of $\eta_F$ in term of $\theta_{F, \Omega(F)}$. Does it seem right ? Does anyone know how to conclude ?
Best Answer
Good so far!
You haven't explicitly written down how $\Gamma$ acts on morphisms, but it is clear that for $g\in\hom_\mathcal{C}(C,D)$, $\Gamma(g): \Gamma(C)\to\Gamma(D)$ is the natural transformation with $\Gamma(g)_Z = g$ for all objects $Z\in\mathcal{Z}$. In particular, $u = \Gamma(u)_Z$. From here, $u\circ(\eta_F)_Z = \Gamma(u)_Z\circ(\eta_F)_Z = (\Gamma(u)\circ\eta_F)_Z$, so showing that $u\circ(\eta_F)_Z = \alpha_Z$ for every $Z\in\mathcal{Z}$ is the same as simply showing that $\Gamma(u)\circ\eta_F = \alpha$. This statement is true by the adjunction identity $\Gamma(g)\circ\eta = \theta(g)$, true for all $g$ and $g = u = \theta^{-1}(\alpha)$ in particular.