Epimorphism in the functor category $[\mathbf{C}^{op}, \mathbf{Set}]$

Question

Let $\mathbf{C}$ be a small category and $\alpha: F \Rightarrow G$ be a morphism in the functor category $Func(\mathbf{C}^{op}, \mathbf{Set})$. How do I prove that $\alpha$ is an epimorphism if and only if for all $A \in \mathbf{C}$ we have that $\alpha_A : FA \rightarrow GA$ is surjetive? Any tips?

Answer 1

Let $\mathcal{A}$ and $\mathcal{B}$ be categories, $\mathcal{F},\mathcal{G}\colon\mathcal{A}\to\mathcal{B}$ be functors, $\alpha\colon\mathcal{F}\to\mathcal{G}$ be a natural transformation.

If $\alpha(a)$ is an epimorphism for every $a\in\text{Obj}(A)$, then $\alpha$ is an epimorphism in $\text{Func}(\mathcal{A},\mathcal{B})$. It is an easy exercise.

If $\alpha$ is an epimorphism in $\text{Func}(\mathcal{A},\mathcal{B})$ and $\mathcal{B}$ is finitely cocomplete, then $\alpha(a)$ is an epimorphism in $\mathcal{B}$ for every $a\in\text{Obj}(\mathcal{A})$. For every object $a\in\text{Obj}(\mathcal{A})$ denote by $\Delta_a\colon\mathbf{1}\to\mathcal{A}$ such functor, that $\Delta_a(0)=a$. Note, that if $\mathcal{B}$ is finitely cocomplete, then the inverse image functor $\mathcal{B}^{\Delta_a}\colon\mathcal{B}^{\mathcal{A}}\to\mathcal{B}^{\mathbf{1}}$ is right exact. Therefore, the evaluation functor $\text{ev}_a\colon\mathcal{B}^{\mathcal{A}}\to\mathcal{B}$, such that $\text{ev}_a(\mathcal{T})=\mathcal{T}(a)$ for every $\mathcal{T}\in\text{Func}(\mathcal{A},\mathcal{B})$, which is isomorphic to $\mathcal{B}^{\Delta_a}$, preserves epimorphisms.

Then it is sufficient to note that $\mathbf{Set}$ is finitely cocomplete.

Of course, the only difficult part of the proof is the statement that the inverse image functor preserves pointwise colimits. You can read about the theory of pointwise limits/colimits in the Mac Lane's CFWM and in the Borceux's handbook.

Answer 2

Here is a direct calculational argument.. \begin{align*} \quad& \text{ each } α_A : F\,A → G\;A \text{ is epic } \\ ≡ & \color{green}{\{\text{ Definition of epic }\}} \\ & \quad ∀ A • ∀ g,h • \quad h ∘ α_A = g ∘ α_A ⇒ h = g \\ ≡ & \color{green}{\{\text{ ⇒ take h,g to be transformation; ⇐ take ε,η to be constantly h,g }\}} \\ &\quad ∀ A • ∀ η,ε • \quad η_A ∘ α_A = ε_A ∘ α_A ⇒ η_A = ε_A \\ ≡ &\color{green}{\{\text{ Composition of natural transformations }\}} \\ &\quad ∀ A • ∀ η,ε • \quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A \\ ≡ & \color{green}{\{\text{ Quantifier Nesting }\}} & \\ & \quad ∀ η,ε • ∀ A • \quad (η ∘ α)_A = (ε ∘ α)_A ⇒ η_A = ε_A &\\ \Rightarrow & \color{green}{\{\text{ Quantifier distributivity }\}} \\ & \quad ∀ η,ε • \left(∀ A •\; (η ∘ α)_A = (ε ∘ α)_A \right) ⇒ \left(∀ A •\; η_A = ε_A\right) \\ ≡ & \color{green}{\{\text{ Extensionality }\}} \\ & \quad ∀ η,ε •\quad η ∘ α = ε ∘ α ⇒ η = ε \\ ≡ & \color{green}{\{\text{ Definition of epic }\}} \\ & \quad α \text{ is epic } \end{align*}

Note that the since in Set, epic is precisely surjective, the desired result follows --more or less.