Proving the adjunction $\text{ev}_0 \dashv r:\mathcal{C}^{\Delta} \to \mathcal{C}$

Question

I recall that $\Delta$ is the category whose objects are of the form $\textbf{n}=\{0,1,...,n\}$ and morphisms are (weakly) order preserving maps.

Let $\mathcal{C}$ be a category, and let $\mathcal{C}^{\Delta}=[\Delta, \mathcal{C}]$ be the functor category of cosimplicial objects in $\mathcal{C}$.

There is a functor $\text{ev}_0:\mathcal{C}^{\Delta} \to \mathcal{C}$ which takes a cosimplicial object $X[-]$ to its value at $0$, $X[0]$.

There is also a functor $r:\mathcal{C} \to \mathcal{C}^{\Delta}$ taking an object $C$ to the constant functor $rC$ such that $rC[n]=C$ for all $n$.

I read the claim that we have an adjunction $$\text{ev}_0 \dashv r$$ and I would like to prove it.

Given a natural transformation $\eta: X[-] \Rightarrow rC$, I can of course send it to the map $\eta_0:X[0]\to C.$

On the other hand, I can consider the diagram $$\cdots\to X[n]\to \cdots \to X[1]\to X[0]$$ wehere each $$\alpha_{n,n-1}:X[n] \to X[n-1]$$ is induced by the surjection $\textbf{n}\to \textbf{n-1}$ sending $n \mapsto n-1$ and $i \mapsto i$ for all $i<n$.

So given a map $f:X[0] \to C,$ I can inductively define $$f_0=f$$ $$f_i=f_{i-1}\alpha_{i,i-1}$$

I think that if I prove this family $\{f_i\}_i$ defines a map of cosimplicial sets, i.e. a natural transformation, I am done. But I don't know how to do that w.r.t. general maps $X[i]\to X[j].$

Answer 1

For each $n$ there's a unique map $!_n : n \to 0$ in $\Delta$. Suppose that $\alpha : X \implies r(c)$ is a natural transformation. Then by naturality at the map $!_n$, the component $\alpha_n$ must be equal to $\alpha_0 \circ X(!_n)$. Thus a natural transformation in $\mathcal{C}^\Delta(X, r(c))$ is completely determined by $\alpha_0$.

On the other hand, if $\alpha_0 : X(0) \to c$ is a morphism in $\mathcal{C}$ then we can lift it to an natural transformation $\alpha : X \implies r(c)$ by defining the component $\alpha_m : X(m) \to c$ to be $\alpha_0 \circ X(!_m)$. This really is a natural transformation because if $f:n \to m$ in $\Delta$ then $\alpha_m \circ X(f) = \alpha_0 \circ X(!_m) \circ X(f) = \alpha_0 \circ X(!_m \circ f) = \alpha_0 \circ X(!_n) = \alpha_n$.