Situation 14.34.1. Let $\mathcal{A}$, $\mathcal{S}$ be categories and let $V : \mathcal{A} \to \mathcal{S}$ be a functor with a left adjoint $U : \mathcal{S} \to \mathcal{A}$.

## 14.34 Standard resolutions

Some of the material in this section can be found in [Appendix 1, Godement] and [I 1.5, cotangent].

In this very general situation we will construct a simplicial object $X$ in the category of functors from $\mathcal{A}$ to $\mathcal{A}$. We suggest looking at the examples presented later on before reading the text of this section.

For the construction we will use the horizontal composition as defined in Categories, Section 4.28. The definition of the adjunction morphisms^{1}

in Categories, Section 4.24 shows that the compositions

are the identity morphisms. Here to define the morphism $\eta \star 1_ V$ we silently identify $V$ with $\text{id}_\mathcal {S} \circ V$ and $1_ V$ stands for $\text{id}_ V : V \to V$. We will use this notation and these relations repeatedly in what follows. For $n \geq 0$ we set

In other words, $X_ n$ is the $(n + 1)$-fold composition of $U \circ V$ with itself. We also set $X_{-1} = \text{id}_\mathcal {A}$. We have $X_{n + m + 1} = X_ n \circ X_ m$ for all $n, m \geq -1$. We will endow this sequence of functors with the structure of a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ by constructing the morphisms of functors

satisfying the relations displayed in Lemma 14.2.3. Namely, we set

Finally, write $\epsilon _0 = d : X_0 \to X_{-1}$.

Lemma 14.34.2. In Situation 14.34.1 the system $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ and $\epsilon _0$ defines an augmentation $\epsilon $ from $X$ to the constant simplicial object with value $X_{-1} = \text{id}_\mathcal {A}$.

**Proof.**
Consider $Y = U \circ V : \mathcal{A} \to \mathcal{A}$. We already have the transformation $d : Y = U \circ V \to \text{id}_\mathcal {A}$. Let us denote

This places us in the sitation of Example 14.33.1. It is immediate from the formulas that the $X, d^ n_ i, s^ n_ i$ constructed above and the $X, s^ n_ i, s^ n_ i$ constructed from $Y, d, s$ in Example 14.33.1 agree. Thus, according to Lemma 14.33.2 it suffices to prove that

The first equal sign translates into the equality

which holds if we have $1_ U = (d \star 1_ U) \circ (1_ U \star \eta )$ which in turn holds by (14.34.1.1). Similarly for the second equal sign. For the last equation we need to prove

For this it suffices to prove $(\eta \star 1_ V \star 1_ U) \circ \eta = (1_ V \star 1_ U \star \eta ) \circ \eta $ which is true because both sides are the same as $\eta \star \eta $. $\square$

Before reading the proof of the following lemma, we advise the reader to look at the example discussed in Example 14.34.8 in order to understand the purpose of the lemma.

Lemma 14.34.3. In Situation 14.34.1 the maps

are homotopy equivalences.

**Proof.**
As in the proof of Lemma 14.34.2 we set $Y = U \circ V$ so that we are in the sitation of Example 14.33.1.

Proof of the first homotopy equivalence. By Lemma 14.33.4 to construct a map $h : V \to V \circ X$ right inverse to $1_ V \star \epsilon $ it suffices to construct a map $h_0 : V \to V \circ Y = V \circ U \circ V$ such that $1_ V = (1_ V \star d) \circ h_0$. Of course we take $h_0 = \eta \star 1_ V$ and the equality holds by (14.34.1.1). To finish the proof we need to show the two maps

are homotopic. This follows immediately from Lemma 14.33.5 (with $G = G' = V$ and $F = F' = \text{id}_\mathcal {S}$).

The proof of the second homotopy equivalence. By Lemma 14.33.4 to construct a map $h : U \to X \circ U$ right inverse to $\epsilon \star 1_ U$ it suffices to construct a map $h_0 : U \to Y \circ U = U \circ V \circ U$ such that $1_ U = (d \star 1_ U) \circ h_0$. Of course we take $h_0 = 1_ U \star \eta $ and the equality holds by (14.34.1.1). To finish the proof we need to show the two maps

are homotopic. This follows immediately from Lemma 14.33.5 (with $G = G' = \text{id}_\mathcal {A}$ and $F = F' = U$). $\square$

Example 14.34.4. Let $R$ be a ring. As an example of the above we can take $i : \text{Mod}_ R \to \textit{Sets}$ to be the forgetful functor and $F : \textit{Sets} \to \text{Mod}_ R$ to be the functor that associates to a set $E$ the free $R$-module $R[E]$ on $E$. For an $R$-module $M$ the simplicial $R$-module $X(M)$ will have the following shape

which comes with an augmentation towards $M$. We will also show this augmentation is a homotopy equivalence of sets. By Lemmas 14.30.8, 14.31.9, and 14.31.8 this is equivalent to asking $M$ to be the only nonzero cohomology group of the chain complex associated to the simplicial module $X(M)$.

Example 14.34.5. Let $A$ be a ring. Let $\textit{Alg}_ A$ be the category of commutative $A$-algebras. As an example of the above we can take $i : \textit{Alg}_ A \to \textit{Sets}$ to be the forgetful functor and $F : \textit{Sets} \to \textit{Alg}_ A$ to be the functor that associates to a set $E$ the polynomial algebra $A[E]$ on $E$ over $A$. (We apologize for the overlap in notation between this example and Example 14.34.4.) For an $A$-algebra $B$ the simplicial $A$-algebra $X(B)$ will have the following shape

which comes with an augmentation towards $B$. We will also show this augmentation is a homotopy equivalence of sets. By Lemmas 14.30.8, 14.31.9, and 14.31.8 this is equivalent to asking $B$ to be the only nonzero cohomology group of the chain complex of $A$-modules associated to $X(B)$ viewed as a simplicial $A$-module.

Example 14.34.6. In Example 14.34.4 we have $X_ n(M) = R[R[\ldots [M]\ldots ]]$ with $n + 1$ brackets. We describe the maps constructed above using a typical element

of $X_1(M)$. The maps $d_0, d_1 : R[R[M]] \to R[M]$ are given by

The maps $s_0, s_1 : R[R[M]] \to R[R[R[M]]]$ are given by

Example 14.34.7. In Example 14.34.5 we have $X_ n(B) = A[A[\ldots [B]\ldots ]]$ with $n + 1$ brackets. We describe the maps constructed above using a typical element

where for each $i, j$ we can write

Obviously this is horrendous! To ease the notation, to see what the $A$-algebra maps $d_0, d_1 : A[A[B]] \to A[B]$ are doing it suffices to see what happens to the variables $[x]$ where

is a general element. For these we get

The maps $s_0, s_1 : A[A[B]] \to A[A[A[B]]]$ are given by

Example 14.34.8. Going back to the example discussed in Example 14.34.5 our Lemma 14.34.3 signifies that for any ring map $A \to B$ the map of simplicial rings

is a homotopy equivalence on underlying simplicial sets. Moreover, the inverse map constructed in Lemma 14.34.3 is in degree $n$ given by

with obvious notation. In the other direction the lemma tells us that for every set $E$ there is a homotopy equivalence

of rings. The inverse map constructed in the lemma is in degree $n$ given by the ring map

(with obvious notation).

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