## 14.28 Homotopies and cosimplicial objects

Let $\mathcal{C}$ be a category with finite products. Let $V$ be a cosimplicial object and consider $\mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)$, see Section 14.14. The morphisms $e_0, e_1 : \Delta [0] \to \Delta [1]$ produce two morphisms $e_0, e_1 : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V) \to V$.

Definition 14.28.1. Let $\mathcal{C}$ be a category having finite products. Let $U$ and $V$ be two cosimplicial objects of $\mathcal{C}$. Let $a, b : U \to V$ be two morphisms of cosimplicial objects of $\mathcal{C}$.

We say a morphism

\[ h : U \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V) \]

such that $a = e_0 \circ h$ and $b = e_1 \circ h$ is a *homotopy from $a$ to $b$*.

We say $a$ and $b$ are *homotopic* or are *in the same homotopy class* if there exists a sequence $a = a_0, a_1, \ldots , a_ n = b$ of morphisms from $U$ to $V$ such that for each $i = 1, \ldots , n$ there either exists a homotopy from $a_ i$ to $a_{i - 1}$ or there exists a homotopy from $a_{i - 1}$ to $a_ i$.

This is dual to the notion we introduced for simplicial objects in Section 14.26. To explain this, consider a homotopy $h : U \to \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)$ from $a$ to $b$ as in the definition. Recall that $\Delta [1]_ n$ is a finite set. The degree $n$ component of $h$ is a morphism

\[ h_ n = (h_{n, \alpha }) : U \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)_ n = \prod \nolimits _{\alpha \in \Delta [1]_ n} V_ n \]

The morphisms $h_{n, \alpha } : U_ n \to V_ n$ of $\mathcal{C}$ have the property that for every morphism $f : [n] \to [m]$ of $\Delta $ we have

14.28.1.1
\begin{equation} \label{simplicial-equation-property-homotopy-cosimplicial} h_{m, \alpha } \circ U(f) = V(f) \circ h_{n, \alpha \circ f} \end{equation}

Moreover, the condition that $a = e_0 \circ h$ means that $a_ n = h_{n, 0 : [n] \to [1]}$ where $0 : [n] \to [1]$ is the constant map with value $0$. Similarly, the condition that $b = e_1 \circ h$ means that $b_ n = h_{n, 1 : [n] \to [1]}$ where $1 : [n] \to [1]$ is the constant map with value $1$. Conversly, given a family of morphisms $\{ h_{n, \alpha }\} $ such that (14.28.1.1) holds for all morphisms $f$ of $\Delta $ and such that $a_ n = h_{n, 0 : [n] \to [1]}$ and $b_ n = h_{n, 1 : [n] \to [1]}$ for all $n \geq 0$, then we obtain a homotopy $h$ from $a$ to $b$ by setting $h = \prod _{\alpha \in \Delta [1]_ n} h_{n, \alpha }$.

Lemma 14.28.3. Let $\mathcal{C}$ be a category. Suppose that $U$ and $V$ are two cosimplicial objects of $\mathcal{C}$. Let $a, b : U \to V$ be morphisms of cosimplicial objects. Recall that $U$, $V$ correspond to simplicial objects $U'$, $V'$ of $\mathcal{C}^{opp}$. Moreover $a, b$ correspond to morphisms $a', b' : V' \to U'$. The following are equivalent

There exists a homotopy $h = \{ h_{n, \alpha }\} $ from $a$ to $b$ as in Remark 14.28.2.

There exists a homotopy $h = \{ h_{n, i}\} $ from $a'$ to $b'$ as in Remark 14.26.4.

Thus $a$ is homotopic to $b$ as in Remark 14.28.2 if and only if $a'$ is homotopic to $b'$ as in Remark 14.26.4.

**Proof.**
In case $\mathcal{C}$ has finite products, then $\mathcal{C}^{opp}$ has finite coproducts and we may use Definitions 14.28.1 and 14.26.1 instead of Remarks 14.28.2 and 14.26.4. In this case $h : U \to \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)$ is the same as a morphism $h' : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)' \to U'$. Since products and coproducts get switched too, it is immediate that $(\mathop{\mathrm{Hom}}\nolimits (\Delta [1], V))' = V' \times \Delta [1]$. Moreover, the “primed” version of the morphisms $e_0, e_1 : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V) \to V$ are the morphisms $e_0, e_1 : V' \to \Delta [1] \times V$. Thus $e_0 \circ h = a$ translates into $h' \circ e_0 = a'$ and similarly $e_1 \circ h = b$ translates into $h' \circ e_1 = b'$. This proves the lemma in this case.

In the general case, one needs to translate the relations given by (14.28.1.1) into the relations given in Lemma 14.26.2. We omit the details.

The final assertion is formal from the equivalence of (1) and (2).
$\square$

slogan
Lemma 14.28.4. Let $\mathcal{C}, \mathcal{C}', \mathcal{D}, \mathcal{D}'$ be categories. With terminology as in Remarks 14.28.2 and 14.26.4.

Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{D}$. Let $F : \mathcal{D} \to \mathcal{D}'$ be a covariant functor. If $a$ and $b$ are homotopic, then $F(a)$, $F(b)$ are homotopic morphisms $F(U) \to F(V)$ of simplicial objects.

Let $a, b : U \to V$ be morphisms of cosimplicial objects of $\mathcal{C}$. Let $F : \mathcal{C} \to \mathcal{C}'$ be a covariant functor. If $a$ and $b$ are homotopic, then $F(a)$, $F(b)$ are homotopic morphisms $F(U) \to F(V)$ of cosimplicial objects.

Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{D}$. Let $F : \mathcal{D} \to \mathcal{C}$ be a contravariant functor. If $a$ and $b$ are homotopic, then $F(a)$, $F(b)$ are homotopic morphisms $F(V) \to F(U)$ of cosimplicial objects.

Let $a, b : U \to V$ be morphisms of cosimplicial objects of $\mathcal{C}$. Let $F : \mathcal{C} \to \mathcal{D}$ be a contravariant functor. If $a$ and $b$ are homotopic, then $F(a)$, $F(b)$ are homotopic morphisms $F(V) \to F(U)$ of simplicial objects.

**Proof.**
By Lemma 14.28.3 above, we can turn $F$ into a covariant functor between a pair of categories, and we have to show that the functor preserves homotopic pairs of maps. This is explained in Remark 14.26.4.
$\square$

Lemma 14.28.5. Let $f : X \to Y$ be a morphism of a category $\mathcal{C}$ with pushouts. Assume there is a morphism $s : Y \to X$ with $s \circ f = \text{id}_ X$. Consider the cosimplicial object $U$ constructed in Example 14.5.5 starting with $f$. The morphism $U \to U$ which in each degree is the self map of $Y \amalg _ X \ldots \amalg _ X Y$ given by $f \circ s$ on each factor is homotopic to the identity on $U$. In particular, $U$ is homotopy equivalent to the constant cosimplicial object $X$.

**Proof.**
This lemma is dual to Lemma 14.26.9. Hence this lemma follows on applying Lemma 14.28.3.
$\square$

slogan
Lemma 14.28.6. Let $\mathcal{A}$ be an additive category. Let $a, b : U \to V$ be morphisms of cosimplicial objects of $\mathcal{A}$. If $a$, $b$ are homotopic, then $s(a), s(b) : s(U) \to s(V)$ are homotopic maps of cochain complexes. If in addition $\mathcal{A}$ is abelian, then $Q(a), Q(b) : Q(U) \to Q(V)$ are homotopic maps of cochain complexes.

**Proof.**
Let $(-)' : \mathcal{A} \to \mathcal{A}^{opp}$ be the contravariant functor $A \mapsto A$. By Lemma 14.28.5 the maps $a'$ and $b'$ are homotopic. By Lemma 14.27.1 we see that $s(a')$ and $s(b')$ are homotopic maps of chain complexes. Since $s(a') = (s(a))'$ and $s(b') = (s(b))'$ we conclude that also $s(a)$ and $s(b)$ are homotopic by applying the additive contravariant functor $(-)'' : \mathcal{A}^{opp} \to \mathcal{A}$. The result for the $Q$-complexes follows in the same manner using that $Q(U)' = N(U')$.
$\square$

Lemma 14.28.7. Let $\mathcal{A}$ be an additive category. Let $a : U \to V$ be a morphism of cosimplicial objects of $\mathcal{A}$. If $a$ is a homotopy equivalence, then $s(a) : s(U) \to s(V)$ is a homotopy equivalence of chain complexes. If in addition $\mathcal{A}$ is abelian, then also $Q(a) : Q(U) \to Q(V)$ is a homotopy equivalence of chain complexes.

**Proof.**
Omitted. See Lemma 14.28.6 above.
$\square$

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