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

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.

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 $$X[i]\to X[j].$$

• You may be interested in this related question.
– jgon
Aug 19, 2020 at 19:53
• The point is that $0$ is terminal in $\Delta$ Aug 19, 2020 at 19:57

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 $$\alpha_m \circ X(f) = \alpha_0 \circ X(!_m) \circ X(f) = \alpha_0 \circ X(!_m \circ f) = \alpha_0 \circ X(!_n) = \alpha_n$$.