Situation 14.33.1. Let $\mathcal{A}$, $\mathcal{S}$ be categories and let $i : \mathcal{A} \to \mathcal{S}$ be a functor with a left adjoint $F : \mathcal{S} \to \mathcal{A}$.

## 14.33 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 $\mathcal{A}$. Please keep the following example in mind while we do this.

Example 14.33.2. As an example of the above we can take $i : \textit{Rings} \to \textit{Sets}$ to be the forgetful functor and $F : \textit{Sets} \to \textit{Rings}$ to be the functor that associates to a set $E$ the polynomial algebra $\mathbf{Z}[E]$ on $E$ over $\mathbf{Z}$. The simplicial object $X$ when evaluated on an ring $A$ will give the simplicial ring

which comes with an augmentation towards $A$. We will also show this augmentation is a homotopy equivalence.

For the general construction we will use the horizontal composition as defined in Categories, Section 4.27. The definition of the adjunction morphisms $k : F \circ i \to \text{id}_\mathcal {A}$ and $t : \text{id}_\mathcal {S} \to i \circ F$ in Categories, Section 4.24 shows that the compositions

are the identity morphisms. Here to define the morphism $t \star 1$ we silently identify $i$ with $\text{id}_\mathcal {S} \circ i$ and $1$ stands for $\text{id}_ i : i \to i$. 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 $F \circ i$ 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 $\epsilon _0 = k : X_0 \to X_{-1}$.

Example 14.33.3. In Example 14.33.2 we have $X_ n(A) = \mathbf{Z}[\mathbf{Z}[\ldots [A]\ldots ]]$ with $n + 1$ brackets. We describe the maps constructed above using a typical element $\xi = \sum n_ i[n_{ij}[a_{ij}]]$ of $X_1(A)$. The maps $d_0, d_1 : \mathbf{Z}[\mathbf{Z}[A]] \to \mathbf{Z}[A]$ are given by

The maps $s_0, s_1 : \mathbf{Z}[\mathbf{Z}[A]] \to \mathbf{Z}[\mathbf{Z}[\mathbf{Z}[A]]]$ are given by

Lemma 14.33.4. In Situation 14.33.1 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.**
Suppose that we have shown that $X$ is a simplicial object. Then to prove that $\epsilon _0 = k$ defines an augmentation we have to check that $\epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1$ as morphisms $X_1 \to X_{-1}$, see Lemma 14.20.2. In other words, we have to check that the diagram

is commutative. More precisely we should write this as the equality

as morphisms $(F \circ i) \circ (F \circ i) \to \text{id}_\mathcal {A} \circ \text{id}_\mathcal {A}$. Applying the general property of Categories, Lemma 4.27.2 both sides expand to $k \star k$ when equality holds.

To prove that $X$ is a simplicial object we have to check (see Remark 14.3.3):

If $i < j$, then $d_ i \circ d_ j = d_{j - 1} \circ d_ i$.

If $i < j$, then $d_ i \circ s_ j = s_{j - 1} \circ d_ i$.

We have $\text{id} = d_ j \circ s_ j = d_{j + 1} \circ s_ j$.

If $i > j + 1$, then $d_ i \circ s_ j = s_ j \circ d_{i - 1}$.

If $i \leq j$, then $s_ i \circ s_ j = s_{j + 1} \circ s_ i$.

Relation (1) is proved in exactly the same manner as the proof of the equality $\epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1$ above.

The simplest case of equality (5) is the commutativity of the diagram

which holds because both compositions expand to the morphism $1_ F \star t \star t \star 1_ i$ from $F \circ \text{id}_\mathcal {A} \circ \text{id}_\mathcal {A} \circ i$ to $F \circ (i \circ F) \circ (i \circ F) \circ i$. All other cases of (5) are proved in the same manner.

The simplest case of equalities (2) and (4) is the commutativity of the diagram

which again holds because both compositions expand to give $1_ F \star k \star 1_ i \star t$ as maps from $F \circ (i \circ F) \circ i \circ \text{id}_\mathcal {A}$ to $F \circ \text{id}_\mathcal {A} \circ i \circ (F \circ i)$. All other cases of (2) and (4) are proved in the same manner.

The relations (3) are the only nontrivial ones and these are consequences of the fact that the compositions in (14.33.2.1) are the identity. For example, the simplest case of (3) states that the compositions

go around the diagram either way evaluate out to the identity. Going around the top the composition evaluates to $1_ F \star ((k \star 1_ i) \circ (1_ i \star t))$ which is the identity by what was said above. The other cases of (3) are proved in the same manner. $\square$

Before reading the proof of the following lemma, we strongly urge the reader to look at the example discussed in Example 14.33.6 in order to understand the purpose of the lemma.

Lemma 14.33.5. In Situation 14.33.1 the maps

are homotopy equivalences.

**Proof.**
Denote $\epsilon _ n : X_ n \to X_{-1}$ the components of the augmentation morphism. We observe that $\epsilon _ n = k^{\star (n + 1)}$, the $(n + 1)$-fold $\star $-composition of $k$. Recall that $t : \text{id}_\mathcal {S} \to i \circ F$ is the adjunction map. We have the morphisms

which are right inverse to $1_ i \star \epsilon _ n$ and the morphisms

which are right inverse to $\epsilon _ n \star 1_ F$. These morphisms determine morphisms of simplicial objects $b : i \to i \circ X$ and $c : F \to X \circ F$ (proof omitted). To finish it suffices to construct a homotopy between the morphisms $1, b \circ (1_ i \star \epsilon ) : i \circ X \to i \circ X$ and between the two morphisms $1, c \circ (\epsilon \star 1_ F) : X \circ F \to X \circ F$.

To show the morphisms $b \circ (1_ i \star \epsilon ), 1 : i \circ X \to i \circ X$ are homotopic we have to construct morphisms

for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and

Thus a logical choice is

Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing

we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal {S}$ using $k^{\star (j)}$, then adding a $\text{id}_\mathcal {S}$ in front and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^ n_ m$ for $j > m$.

We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j} \circ d^ n_ m$ for $j \leq m$.

We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^ n_ m$ for $j > m$.

We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j} \circ s^ n_ m$ for $j \leq m$.

Recall that $d^ n_ m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star $-commute with $k$, but not with $t$). Similarly, $s^ n_ m$ is given by applying $1_ F \star t \star i_ i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted.

To show the morphisms $1, c \circ (\epsilon \star 1_ F) : X \circ F \to X \circ F$ are homotopic we have to construct morphisms

for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and

Thus a logical choice is

Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing

we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal {S}$ using $k^{\star (j)}$, then inserting a $\text{id}_\mathcal {S}$ just after the first $F$ and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^ n_ m$ for $j > m$.

We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j} \circ d^ n_ m$ for $j \leq m$.

We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^ n_ m$ for $j > m$.

We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j} \circ s^ n_ m$ for $j \leq m$.

Recall that $d^ n_ m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star $-commute with $k$, but not with $t$). Similarly, $s^ n_ m$ is given by applying $1_ F \star t \star i_ i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted. $\square$

Example 14.33.6. Going back to the example discussed in Example 14.33.2 our Lemma 14.33.5 signifies that for any ring $A$ the map of simplicial rings

is a homotopy equivalence on underlying simplicial sets. Moreover, the inverse map constructed in Lemma 14.33.5 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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