counit is a mono Let $\mathscr C$ and $\mathscr D$ be categories.  Let $\Delta_{-}$ be the functor $\mathscr D\to   \mathscr D^ \mathscr C$ sending $D$ to the constant functor $\Delta _D$.  I want to prove that if $\Delta_{-}$ has a right adjoint $G$, then $\mathscr D$ has a limiting cone for any functor $\mathscr C \to \mathscr D$.  Assuming $G$ exists I can prove that for any $\Phi \in  \mathscr D^ \mathscr C$, any natural $\nu: \Delta_D \implies \Phi$ will factor through $$\Delta_D \implies\Delta_{G(\Phi)}  \implies \Phi $$ where the second natural transformation is $\epsilon_ \Phi$, the counit for the adjunction.  But I need to show the first arrow is unique, so it seems like I need to show $\epsilon$ is a monomorphism. Now $\Delta_{-}$ is full and faithful which i think implies the unit is an isomorphism, but I don't see any information about $G$ that would help me with $\epsilon$.
 A: Recall.
An adjunction $F ⊣ G$ between functors $F \colon \mathscr{C} \to \mathscr{D}$ and $G \colon \mathscr{D} \to \mathscr{C}$ is given by natural bijections
$$
  H_{X, Y} \colon \mathscr{C}(X, G(Y)) \longrightarrow \mathscr{D}(F(X), Y) \,.
$$
In terms of the counit $ε$ of this adjunction, these bijections are given by the formula
$$
  H_{X, Y}(f) = ε_Y ∘ F(f) \,.
$$

Note first that in the given situation we don’t need a unique natural transformations $μ \colon Δ_D \Rightarrow Δ_{G(Φ)}$ with $ν = ε_Φ ∘ μ$.
Instead, we need a unique morphism $f \colon D \to G(Φ)$ with $ν = ε_Φ ∘ Δ_f$.
The adjunction $Δ_{(-)} ⊣ G$ is given by natural bijections
$$
  H_{D, Φ}
  \colon
  \mathscr{D}(D, G(Φ))
  \longrightarrow
  \mathscr{D}^{\mathscr{C}}(Δ_D, Φ) \,,
$$
that are given in terms of the counit $ε$ by the formula
$$
  H_{D, Φ}(f) = ε_Φ ∘ Δ_f \,.
$$
This tells us that for every $ν ∈ \mathscr{D}^{\mathscr{C}}(Δ_D, Φ)$ there exists a unique $f ∈ \mathscr{D}(D, G(Φ))$ such that $ν = H_{D, Φ}(f)$.
In other words, for every natural transformation $ν \colon Δ_D \Rightarrow Φ$ there exists a unique morphism $f \colon D \to G(Φ)$ such such that $ν$ factors as $ν = ε_Φ ∘ Δ_f$.

We didn’t need $ε_Y$ to be a monomorphism.
However, we have nevertheless the following general results for an adjunction $F ⊣ G$ with unit $η$ and counit $ε$:

*

*The left adjoint $F$ is faithful if and only if the unit $η$ is a monomorphism in each component.


*The right adjoint $G$ is faithful if and only if the counit $ε$ is an epimorphism in each component.


*The left adjoint $F$ is full if and only if the unit $η$ is a split epimorphism in each component.


*The right adjoint $G$ is full if and only if the counit $ε$ is a split monomorphism in each component.
This can be found in Mac Lane’s Categories for the Working Mathematician, Chapter IV, section 3, Theorem 1 (end of page 90).
A: You are on the right track, but we don't need to show that $\epsilon$ is a mono. Here is how I would reason.
Suppose $\Delta ⊣ G$ with unit $\eta$ and counit $\epsilon$. Let $\phi : C \to D$. As you have already noted, we should have that $G(\phi)$ is the limit of $\phi$ with limit cone $\epsilon_\phi : \Delta \circ G (\phi) \Rightarrow \phi$. To prove this, let $\alpha : \Delta (d) \Rightarrow \phi$. So we have:

where we need to show there exists a unique $f$ as in the dotted lines. But, a common lemma regarding adjunctions states:

each component of the (co)unit is a universal arrow.

Specific to this case, we have that each component of $\epsilon$, so in particular $\epsilon_\phi$, is a universal arrow from $\Delta$ to $\phi$. This proves the claim.
