## 14.33 Preparation for standard resolutions

The material in this section can be found in [Appendix 1, Godement]

Example 14.33.1. Let $Y : \mathcal{C} \to \mathcal{C}$ be a functor from a category to itself and suppose given transformations of functors

$d : Y \longrightarrow \text{id}_\mathcal {C} \quad \text{and}\quad s : Y \longrightarrow Y \circ Y$

Using these transformations we can construct something that looks like a simplicial object. Namely, for $n \geq 0$ we define

$X_ n = Y \circ \ldots \circ Y \quad (n + 1\text{ compositions})$

Observe that $X_{n + m + 1} = X_ n \circ X_ m$ for $n, m \geq 0$. Next, for $n \geq 0$ and $0 \leq j \leq n$ we define using notation as in Categories, Section 4.28

$d^ n_ j = 1_{X_{j - 1}} \star d \star 1_{X_{n - j - 1}} : X_ n \to X_{n - 1} \quad \text{and}\quad s^ n_ j = 1_{X_{j - 1}} \star s \star 1_{X_{n - j - 1}} : X_ n \to X_{n + 1}$

So $d^ n_ j$, resp. $s^ n_ j$ is the natural transformation using $d$, resp. $s$ on the $j$th $Y$ (counted from the left) in the composition defining $X_ n$.

Lemma 14.33.2. In Example 14.33.1 if

$1_ Y = (d \star 1_ Y) \circ s = (1_ Y \star d) \circ s \quad \text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s$

then $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object in the category of endofunctors of $\mathcal{C}$ and $d : X_0 = Y \to \text{id}_\mathcal {C}$ defines an augmentation.

Proof. To see that we obtain a simplicial object we have to check that the relations (1)(a) – (e) of Lemma 14.3.2 are satisfied. We will use the short hand notation

$1_ a = 1_{X_{a - 1}} = 1_ Y \star \ldots \star 1_ Y \quad (a\text{ factors})$

for $a \geq 0$. With this notation we have

$d^ n_ j = 1_ j \star d \star 1_{n - j} \quad \text{and}\quad s^ n_ j = 1_ j \star s \star 1_{n - j}$

We are repeatedly going to use the rule that for transformations of funtors $a, a', b, b'$ we have $(a' \circ a) \star (b' \circ b) = (a' \star b') \circ (a \star b)$ provided that the $\star$ and $\circ$ compositions in this formula make sense, see Categories, Lemma 4.28.2.

Condition (1)(a) always holds (no conditions needed on $d$ and $s$). Namely, let $0 \leq i < j \leq n + 1$. We have to show that $d^ n_ i \circ d^{n + 1}_ j = d^ n_{j - 1} \circ d^{n + 1}_ i$, i.e.,

$(1_ i \star d \star 1_{n - i}) \circ (1_ j \star d \star 1_{n + 1 - j}) = (1_{j - 1} \star d \star 1_{n + 1 - j}) \circ (1_ i \star d \star 1_{n + 1 - i})$

We can rewrite the left hand side as

\begin{align*} & (1_ i \star d \star 1_{j - i - 1} \star 1_{n + 1 - j}) \circ (1_ i \star 1_1 \star 1_{j - i - 1} \star d \star 1_{n + 1 - j}) \\ & = 1_ i \star \left((d \star 1_{j - i - 1}) \circ (1_1 \star 1_{j - i - 1} \star d)\right) \star 1_{n + 1 - j} \\ & = 1_ i \star d \star 1_{j - i - 1} \star d \star 1_{n + 1 - j} \end{align*}

The second equality is true because $d \circ 1_1 = d$ and $1_{j - i} \circ (1_{j - i - 1} \star d) = 1_{j - i - 1} \star d$. A similar computation gives the same result for the right hand side.

We check condition (1)(b). Let $0 \leq i < j \leq n - 1$. We have to show that $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i$, i.e.,

$(1_ i \star d \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j - 1} \star s \star 1_{n - 1 - j}) \circ (1_ i \star d \star 1_{n - 1 - i})$

By the same kind of calculus as in case (1)(a) both sides simplify to $1_ i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}$.

We check condition (1)(c). Let $0 \leq j \leq n - 1$. We have to show $\text{id} = d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j$, i.e.,

$1_ n = (1_ j \star d \star 1_{n - j}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star d \star 1_{n - j - 1}) \circ (1_ j \star s \star 1_{n - 1 - j})$

This is easily seen to be implied by the first assumption of the lemma.

We check condition (1)(d). Let $0 < j + 1 < i \leq n$. We have to show $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}$, i.e.,

$(1_ i \star d \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_ j \star s \star 1_{n - 2 - j}) \circ (1_{i - 1} \star d \star 1_{n - i})$

By the same kind of calculus as in case (1)(a) both sides simplify to $1_ j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}$.

We check condition (1)(e). Let $0 \leq i \leq j \leq n - 1$. We have to show that $s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i$, i.e.,

$(1_ i \star s \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star s \star 1_{n - 1 - j}) \circ (1_ i \star s \star 1_{n - 1 - i})$

By the same kind of calculus as in case (1)(a) this reduces to

$(s \star 1_{j - i + 1}) \circ (1_{j - i} \star s) = (1_{j - i + 1} \star s) \circ (s \star 1_{j - i})$

If $j = i$ this is exactly one of the two assumptions of the lemma. For $j > i$ left and right hand side both reduce to the equality $s \star 1_{j - i - 1} \star s$ by calculations similar to those we did in case (1)(a).

Finally, in order to show that $d$ defines an augmentation we have to show that $d \circ (1_1 \star d) = d \circ (d \star 1_1)$ which is true because both sides are equal to $d \star d$. $\square$

Example 14.33.3. Let $\mathcal{C}$, $Y$, $d$, $s$ be as in Example 14.33.1 satisfying the equations of Lemma 14.33.2. Given functors $F : \mathcal{A} \to \mathcal{C}$ and $G : \mathcal{C} \to \mathcal{B}$ we obtain a simplicial object $G \circ X \circ F$ in the category of functors from $\mathcal{A}$ to $\mathcal{B}$ which comes with an augmentation to $G \circ F$.

Lemma 14.33.4. Let $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $Y$, $d$, $s$, $F$, $G$ be as in Example 14.33.3. Given a transformation of functors $h_0 : G \circ F \to G \circ Y \circ F$ such that

$1_{G \circ F} = (1_ G \star d \star 1_ F) \circ h_0$

Then there is a morphism $h : G \circ F \to G \circ X \circ F$ of simplicial objects such that $\epsilon \circ h = \text{id}$ where $\epsilon : G \circ X \circ F \to G \circ F$ is the augmentation.

Proof. Denote $u_ n : Y = X_0 \to X_ n$ the map of the simplicial object $X$ corresponding to the unique morphism $[n] \to [0]$ in $\Delta$. Set $h_ n : G \circ F \to G \circ X_ n \circ F$ equal to $(1_ G \star u_ n \star 1_ F) \circ h_0$.

For any simplicial object $X = (X_ n)$ in any category $u =(u_ n) : X_0 \to X$ is a morphism from the constant simplicial object on $X_0$ to $X$. Hence $h$ is a morphism of simplicial objects because it is the composition of $1_ G \star u \star 1_ F$ and $h_0$.

Let us check that $\epsilon \circ h = \text{id}$. We compute

$\epsilon _ n \circ (1_ G \star u_ n \star 1_ F) \circ h_0 = \epsilon _0 \circ h_0 = \text{id}$

The first equality because $\epsilon$ is a morphism of simplicial objects and the second equality because $\epsilon _0 = (1_ G \star d \star 1_ F)$ and we can apply the assumption in the statement of the lemma. $\square$

Lemma 14.33.5. Let $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $Y$, $d$, $s$, $F$, $G$ be as in Example 14.33.3. Let $F' : \mathcal{A} \to \mathcal{C}$ and $G' : \mathcal{C} \to \mathcal{B}$ be two functors. Let $(a_ n) : G \circ X \to G' \circ X$ be a morphism of simplicial objects compatible via augmentations with $a : G \to G'$. Let $(b_ n) : X \circ F \to X \circ F'$ be a morphism of simplicial objects compatible via augmentations with $b : F \to F'$. Then the two maps

$a \star (b_ n), (a_ n) \star b : G \circ X \circ F \to G' \circ X \circ F'$

are homotopic.

Proof. To show the morphisms are homotopic we construct morphisms

$h_{n, i} : G \circ X_ n \circ F \to G' \circ X_ n \circ F'$

for $n \geq 0$ and $0 \leq i \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. To satisfy condition (1) of Lemma 14.26.2 we are forced to set $h_{n, 0} = a \star b_ n$ and $h_{n , n + 1} = a_ n \star b$. Thus a logical choice is

$h_{n , i} = a_{i - 1} \star b_{n - i}$

for $1 \leq i \leq n$. Setting $a = a_{-1}$ and $b = b_{-1}$ we see the displayed formular holds for $0 \leq i \leq n + 1$.

Recall that

$d^ n_ j = 1_ G \star 1_ j \star d \star 1_{n - j} \star 1_ F$

on $G \circ X \circ F$ where we use the notation $1_ a = 1_{Y \circ \ldots \circ Y}$ introduced in the proof of Lemma 14.33.2. We are going to use below that we can rewrite this as

\begin{align*} d^ n_ j & = d^ j_ j \star 1_{n - j} = d^{j + 1}_ j \star 1_{n - j} = \ldots = d^{n - 1}_ j \star 1_1 \\ & = 1_ j \star d^{n - j}_0 = 1_{j - 1} \star d^{n - j + 1}_1 = \ldots = 1_1 \star d^{n - 1}_{j - 1} \end{align*}

Of course we have the analogous formulae for $d^ n_ j$ on $G' \circ X \circ F'$.

We check condition (2) of Lemma 14.26.2. Let $i > j$. We have to show

$d^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 2} \star b_{n - i}) \circ d^ n_ j$

Since $i - 1 \geq j$ we can use one of the possible descriptions of $d^ n_ j$ to rewrite the left hand side as

$(d^{i - 1}_ j \star 1_{n - i + 1}) \circ (a_{i - 1} \star b_{n - i}) = (d^{i - 1}_ j \circ a_{i - 1}) \star b_{n - i} = (a_{i - 2} \circ d^{i - 1}_ j) \star b_{n - i}$

Similarly the right hand side becomes

$(a_{i - 2} \star b_{n - i}) \circ (d^{i - 1}_ j \star 1_{n - i + 1}) = (a_{i - 2} \circ d^{i - 1}_ j) \star b_{n - i}$

Thus we obtain the same result and (2) is checked.

We check condition (3) of Lemma 14.26.2. Let $i \leq j$. We have to show

$d^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n - 1 - i}) \circ d^ n_ j$

Since $j \geq i$ we may rewrite the left hand side as

$(1_ i \star d^{n - i}_{j - i}) \circ (a_{i - 1} \star b_{n - i}) = a_{i - 1} \star (b_{n - 1 - i} \circ d^{n - i}_{j - i})$

A similar manipulation shows this agrees with the right hand side.

Recall that

$s^ n_ j = 1_ G \star 1_ j \star s \star 1_{n - j} \star 1_ F$

on $G \circ X \circ F$. We are going to use below that we can rewrite this as

\begin{align*} s^ n_ j & = s^ j_ j \star 1_{n - j} = s^{j + 1}_ j \star 1_{n - j - 1} = \ldots = s^{n - 1}_ j \star 1_1 \\ & = 1_ j \star s^{n - j}_0 = 1_{j - 1} \star s^{n - j + 1}_1 = \ldots = 1_1 \star s^{n - 1}_{j - 1} \end{align*}

Of course we have the analogous formulae for $s^ n_ j$ on $G' \circ X \circ F'$.

We check condition (4) of Lemma 14.26.2. Let $i > j$. We have to show

$s^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_ i \star b_{n - i}) \circ s^ n_ j$

Since $i - 1 \geq j$ we can rewrite the left hand side as

$(s^{i - 1}_ j \star 1_{n - i + 1}) \circ (a_{i - 1} \star b_{n - i}) = (s^{i - 1}_ j \circ a_{i - 1}) \star b_{n - i} = (a_ i \circ s^{i - 1}_ j) \star b_{n - i}$

Similarly the right hand side becomes

$(a_ i \star b_{n - i}) \circ (s^{i - 1}_ j \star 1_{n - i + 1}) = (a_ i \circ s^{i - 1}_ j) \star b_{n - i}$

as desired.

We check condition (5) of Lemma 14.26.2. Let $i \leq j$. We have to show

$s^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n + 1 - i}) \circ s^ n_ j$

This equality holds because both sides evaluate to $a_{i - 1} \star (s^{n - i}_{j - i} \circ b_{n - i}) = a_{i - 1} \star (b_{n + 1 - i} \circ s^{n - i}_{j - i})$ by exactly the same arguments as above. $\square$

Lemma 14.33.6. Let $\mathcal{C}$, $Y$, $d$, $s$ be as in Example 14.33.1 satisfying the equations of Lemma 14.33.2. Let $f : \text{id}_\mathcal {C} \to \text{id}_\mathcal {C}$ be an endomorphism of the identity functor. Then $f \star 1_ X, 1_ X \star f : X \to X$ are maps of simplicial objects compatible with $f$ via the augmentation $\epsilon : X \to \text{id}_\mathcal {C}$. Moreover, $f \star 1_ X$ and $1_ X \star f$ are homotopic.

Proof. The map $f \star 1_ X$ is the map with components

$X_ n = \text{id}_\mathcal {C} \circ X_ n \xrightarrow {f \star 1_{X_ n}} \text{id}_\mathcal {C} \circ X_ n = X_ n$

For a transformation $a : F \to G$ of endofunctors of $\mathcal{C}$ we have $a \circ (f \star 1_ F) = f \star a = (f \star 1_ G) \circ a$. Thus $f \star 1_ X$ is indeed a morphism of simplicial objects. Similarly for $1_ X \star f$.

To show the morphisms are homotopic we construct morphisms $h_{n, i} : X_ n \to X_ n$ for $n \geq 0$ and $0 \leq i \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. It turns out we can take

$h_{n, i} = 1_ i \star f \star 1_{n + 1 - i}$

where $1_ i$ is the identity transformation on $Y \circ \ldots \circ Y$ as in the proof of Lemma 14.33.2. We have $h_{n, 0} = f \star 1_{X_ n}$ and $h_{n, n + 1} = 1_{X_ n} \star f$ which checks the first condition. In checking the other conditions we use the comments made in the proof of Lemma 14.33.5 about the maps $d^ n_ j$ and $s^ n_ j$.

We check condition (2) of Lemma 14.26.2. Let $i > j$. We have to show

$d^ n_ j \circ (1_ i \star f \star 1_{n + 1 - i}) = (1_{i - 1} \star f \star 1_{n + 1 - i}) \circ d^ n_ j$

Since $i - 1 \geq j$ we can use one of the possible descriptions of $d^ n_ j$ to rewrite the left hand side as

$(d^{i - 1}_ j \star 1_{n - i + 1}) \circ (1_ i \star f \star 1_{n + 1 - i}) = d^{i - 1}_ j \star f \star 1_{n + 1 - i}$

Similarly the right hand side becomes

$(1_{i - 1} \star f \star 1_{n + 1 - i}) \circ (d^{i - 1}_ j \star 1_{n - i + 1}) = d^{i - 1}_ j \star f \star 1_{n + 1 - i}$

Thus we obtain the same result and (2) is checked.

The conditions (3), (4), and (5) of Lemma 14.26.2 are checked in exactly the same manner using the strategy of the proof of Lemma 14.33.5. We omit the details1. $\square$

[1] When $f$ is invertible it suffices to prove that $(a_ n) = 1_ X$ and $(b_ n) = f^{-1} \star 1_ X \star f$ are homotopic. But this follows from Lemma 14.33.5 because in this case $a = b = 1_{\text{id}_\mathcal {C}}$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).