1
$\begingroup$

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 ?

$\endgroup$
2
  • $\begingroup$ Crossposted on MSE. $\endgroup$
    – Zhen Lin
    Jan 21, 2021 at 22:20
  • 2
    $\begingroup$ Use that $\theta_{F,C}$ is natural in $F$ and $C$. As you said, every natural transformation $\alpha$ on the right gives a cone under $F$. The cones are in bijection with morphisms $f_\alpha$ out of $\Omega(F)$ on the left. For each $\alpha$ we have a commutative diagram $$\begin{matrix}\text{Hom}_\mathcal{C}(\Omega(F), \Omega(F)) & \simeq &\text{Nat}(F, \Gamma(\Omega(F)))\\ \downarrow f_\alpha\circ -& & \downarrow \\ \text{Hom}_\mathcal{C}(\Omega(F), C)& \simeq &\text{Nat}(F, \Gamma(C))\end{matrix}$$ by naturality, so the cone you started with factors through $f_\alpha$ as you want. $\endgroup$
    – Nati PT
    Jan 21, 2021 at 22:53

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.