The Stacks project

14.26 Homotopies

Consider the simplicial sets $\Delta [0]$ and $\Delta [1]$. Recall that there are two morphisms

\[ e_0, e_1 : \Delta [0] \longrightarrow \Delta [1], \]

coming from the morphisms $[0] \to [1]$ mapping $0$ to an element of $[1] = \{ 0, 1\} $. Recall also that each set $\Delta [1]_ k$ is finite. Hence, if the category $\mathcal{C}$ has finite coproducts, then we can form the product

\[ U \times \Delta [1] \]

for any simplicial object $U$ of $\mathcal{C}$, see Definition 14.13.1. Note that $\Delta [0]$ has the property that $\Delta [0]_ k = \{ *\} $ is a singleton for all $k \geq 0$. Hence $U \times \Delta [0] = U$. Thus $e_0, e_1$ above gives rise to morphisms

\[ e_0, e_1 : U \to U \times \Delta [1]. \]

Definition 14.26.1. Let $\mathcal{C}$ be a category having finite coproducts. Suppose that $U$ and $V$ are two simplicial objects of $\mathcal{C}$. Let $a, b : U \to V$ be two morphisms.

  1. We say a morphism

    \[ h : U \times \Delta [1] \longrightarrow V \]

    is a homotopy from $a$ to $b$ if $a = h \circ e_0$ and $b = h \circ e_1$.

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

The relation “there is a homotopy from $a$ to $b$” is in general not transitive or symmetric; we will see it is reflexive in Example 14.26.3. Of course, “being homotopic” is an equivalence relation on the set $\mathop{\mathrm{Mor}}\nolimits (U, V)$ and it is the equivalence relation generated by the relation “there is a homotopy from $a$ to $b$” . It turns out we can define homotopies between pairs of maps of simplicial objects in any category. We will do this in Remark 14.26.4 after we work out in some detail what it means to have a morphism $h : U \times \Delta [1] \to V$.

Let $\mathcal{C}$ be a category with finite coproducts. Let $U$, $V$ be simplicial objects of $\mathcal{C}$. Let $a, b : U \to V$ be morphisms. Further, suppose that $h : U \times \Delta [1] \to V$ is a homotopy from $a$ to $b$. For every $n \geq 0$ let us write

\[ \Delta [1]_ n = \{ \alpha ^ n_0, \ldots , \alpha ^ n_{n + 1}\} \]

where $\alpha ^ n_ i : [n] \to [1]$ is the map such that

\[ \alpha ^ n_ i(j) = \left\{ \begin{matrix} 0 & \text{if} & j < i \\ 1 & \text{if} & j \geq i \end{matrix} \right. \]

Thus

\[ h_ n : (U \times \Delta [1])_ n = \coprod U_ n \cdot \alpha ^ n_ i \longrightarrow V_ n \]

has a component $h_{n, i} : U_ n \to V_ n$ which is the restriction to the summand corresponding to $\alpha ^ n_ i$ for all $i = 0, \ldots , n + 1$.

Lemma 14.26.2. In the situation above, we have the following relations:

  1. We have $h_{n, 0} = b_ n$ and $h_{n, n + 1} = a_ n$.

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

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

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

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

Conversely, given a system of maps $h_{n, i}$ satisfying the properties listed above, then these define a morphism $h$ which is a homotopy from $a$ to $b$.

Proof. Omitted. You can prove the last statement using the fact, see Lemma 14.2.4 that to give a morphism of simplicial objects is the same as giving a sequence of morphisms $h_ n$ commuting with all $d^ n_ j$ and $s^ n_ j$. $\square$

Example 14.26.3. Suppose in the situation above $a = b$. Then there is a trivial homotopy from $a$ to $b$, namely the one with $h_{n, i} = a_ n = b_ n$.

Remark 14.26.4. Let $\mathcal{C}$ be any category (no assumptions whatsoever). Let $U$ and $V$ be simplicial objects of $\mathcal{C}$. Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{C}$. A homotopy from $a$ to $b$ is given by morphisms1 $h_{n, i} : U_ n \to V_ n$, for $n \geq 0$, $i = 0, \ldots , n + 1$ satisfying the relations of Lemma 14.26.2. As in Definition 14.26.1 we say the morphisms $a$ and $b$ are homotopic if there exists a sequence of morphisms $a = a_0, a_1, \ldots , a_ n = b$ from $U$ to $V$ such that for each $i = 1, \ldots , n$ there either exists a homotopy from $a_{i - 1}$ to $a_ i$ or there exists a homotopy from $a_ i$ to $a_{i - 1}$. Clearly, if $F : \mathcal{C} \to \mathcal{C}'$ is any functor and $\{ h_{n, i}\} $ is a homotopy from $a$ to $b$, then $\{ F(h_{n, i})\} $ is a homotopy from $F(a)$ to $F(b)$. Similarly, if $a$ and $b$ are homotopic, then $F(a)$ and $F(b)$ are homotopic. Since the lemma says that the newer notion is the same as the old one in case finite coproduct exist, we deduce in particular that functors preserve the original notion whenever both categories have finite coproducts.

Remark 14.26.5. Let $\mathcal{C}$ be any category. Suppose two morphisms $a, a' : U \to V$ of simplicial objects are homotopic. Then for any morphism $b : V \to W$ the two maps $b \circ a, b \circ a' : U \to W$ are homotopic. Similarly, for any morphism $c : X \to U$ the two maps $a \circ c, a' \circ c : X \to V$ are homotopic. In fact the maps $b \circ a \circ c, b \circ a' \circ c : X \to W$ are homotopic. Namely, if the maps $h_{n, i} : U_ n \to V_ n$ define a homotopy from $a$ to $a'$ then the maps $b \circ h_{n, i} \circ c$ define a homotopy from $b \circ a \circ c$ to $b \circ a' \circ c$. In this way we see that we obtain a new category $\text{hSimp}(\mathcal{C})$ with the same objects as $\text{Simp}(\mathcal{C})$ but whose morphisms are homotopy classes of of morphisms of $\text{Simp}(\mathcal{C})$. Thus there is a canonical functor

\[ \text{Simp}(\mathcal{C}) \longrightarrow \text{hSimp}(\mathcal{C}) \]

which is essentially surjective and surjective on sets of morphisms.

Definition 14.26.6. Let $U$ and $V$ be two simplicial objects of a category $\mathcal{C}$. We say a morphism $a : U \to V$ is a homotopy equivalence if there exists a morphism $b : V \to U$ such that $a \circ b$ is homotopic to $\text{id}_ V$ and $b \circ a$ is homotopic to $\text{id}_ U$. We say $U$ and $V$ are homotopy equivalent if there exists a homotopy equivalence $a : U \to V$.

Example 14.26.7. The simplicial set $\Delta [m]$ is homotopy equivalent to $\Delta [0]$. Namely, consider the unique morphism $f : \Delta [m] \to \Delta [0]$ and the morphism $g : \Delta [0] \to \Delta [m]$ given by the inclusion of the last $0$-simplex of $\Delta [m]$. We have $f \circ g = \text{id}$. We will give a homotopy $h : \Delta [m] \times \Delta [1] \to \Delta [m]$ from $\text{id}_{\Delta [m]}$ to $g \circ f$. Namely $h$ is given by the maps

\[ \mathop{\mathrm{Mor}}\nolimits _\Delta ([n], [m]) \times \mathop{\mathrm{Mor}}\nolimits _\Delta ([n], [1]) \to \mathop{\mathrm{Mor}}\nolimits _\Delta ([n], [m]) \]

which send $(\varphi , \alpha )$ to

\[ k \mapsto \left\{ \begin{matrix} \varphi (k) & \text{if} & \alpha (k) = 0 \\ m & \text{if} & \alpha (k) = 1 \end{matrix} \right. \]

Note that this only works because we took $g$ to be the inclusion of the last $0$-simplex. If we took $g$ to be the inclusion of the first $0$-simplex we could find a homotopy from $g \circ f$ to $\text{id}_{\Delta [m]}$. This is an illustration of the asymmetry inherent in homotopies in the category of simplicial sets.

The following lemma says that $U \times \Delta [1]$ is homotopy equivalent to $U$.

Lemma 14.26.8. Let $\mathcal{C}$ be a category with finite coproducts. Let $U$ be a simplicial object of $\mathcal{C}$. Consider the maps $e_1, e_0 : U \to U \times \Delta [1]$, and $\pi : U \times \Delta [1] \to U$, see Lemma 14.13.3.

  1. We have $\pi \circ e_1 = \pi \circ e_0 = \text{id}_ U$, and

  2. The morphisms $\text{id}_{U \times \Delta [1]}$, and $e_0 \circ \pi $ are homotopic.

  3. The morphisms $\text{id}_{U \times \Delta [1]}$, and $e_1 \circ \pi $ are homotopic.

Proof. The first assertion is trivial. For the second, consider the map of simplicial sets $\Delta [1] \times \Delta [1] \longrightarrow \Delta [1]$ which in degree $n$ assigns to a pair $(\beta _1, \beta _2)$, $\beta _ i : [n] \to [1]$ the morphism $\beta : [n] \to [1]$ defined by the rule

\[ \beta (i) = \max \{ \beta _1(i), \beta _2(i)\} . \]

It is a morphism of simplicial sets, because the action $\Delta [1](\varphi ) : \Delta [1]_ n \to \Delta [1]_ m$ of $\varphi : [m] \to [n]$ is by precomposing. Clearly, using notation from Section 14.26, we have $\beta = \beta _1$ if $\beta _2 = \alpha ^ n_0$ and $\beta = \alpha ^ n_{n + 1}$ if $\beta _2 = \alpha ^ n_{n + 1}$. This implies easily that the induced morphism

\[ U \times \Delta [1] \times \Delta [1] \longrightarrow U \times \Delta [1] \]

of Lemma 14.13.3 is a homotopy from $\text{id}_{U \times \Delta [1]}$ to $e_0 \circ \pi $. Similarly for $e_1 \circ \pi $ (use minimum instead of maximum). $\square$

Lemma 14.26.9. Let $f : Y \to X$ be a morphism of a category $\mathcal{C}$ with fibre products. Assume $f$ has a section $s$. Consider the simplicial object $U$ constructed in Example 14.3.5 starting with $f$. The morphism $U \to U$ which in each degree is the self map $(s \circ f)^{n + 1}$ of $Y \times _ X \ldots \times _ X Y$ given by $s \circ f$ on each factor is homotopic to the identity on $U$. In particular, $U$ is homotopy equivalent to the constant simplicial object $X$.

Proof. Set $g^0 = \text{id}_ Y$ and $g^1 = s \circ f$. We use the morphisms

\begin{eqnarray*} Y \times _ X \ldots \times _ X Y \times \mathop{\mathrm{Mor}}\nolimits ([n], [1]) & \to & Y \times _ X \ldots \times _ X Y \\ (y_0, \ldots , y_ n) \times \alpha & \mapsto & (g^{\alpha (0)}(y_0), \ldots , g^{\alpha (n)}(y_ n)) \end{eqnarray*}

where we use the functor of points point of view to define the maps. Another way to say this is to say that $h_{n, 0} = \text{id}$, $h_{n, n + 1} = (s \circ f)^{n + 1}$ and $h_{n, i} = \text{id}_ Y^{i + 1} \times (s \circ f)^{n + 1 - i}$. We leave it to the reader to show that these satisfy the relations of Lemma 14.26.2. Hence they define the desired homotopy. See also Remark 14.26.4 which shows that we do not need to assume anything else on the category $\mathcal{C}$. $\square$

Lemma 14.26.10. Let $\mathcal{C}$ be a category. Let $T$ be a set. For $t \in T$ let $X_ t$, $Y_ t$ be simplicial objects of $\mathcal{C}$. Assume $X = \prod _{t \in T} X_ t$ and $Y = \prod _{t \in T} Y_ t$ exist.

  1. If $X_ t$ and $Y_ t$ are homotopy equivalent for all $t \in T$ and $T$ is finite, then $X$ and $Y$ are homotopy equivalent.

For $t \in T$ let $a_ t, b_ t : X_ t \to Y_ t$ be morphisms. Set $a = \prod a_ t : X \to Y$ and $b = \prod b_ t : X \to Y$.

  1. If there exists a homotopy from $a_ t$ to $b_ t$ for all $t \in T$, then there exists a homotopy from $a$ to $b$.

  2. If $T$ is finite and $a_ t, b_ t : X_ t \to Y_ t$ for $t \in T$ are homotopic, then $a$ and $b$ are homotopic.

Proof. If $h_ t = (h_{t, n , i})$ is a homotopy from $a_ t$ to $b_ t$ (see Remark 14.26.4), then $h = (\prod _ t h_{t, n, i})$ is a homotopy from $\prod a_ t$ to $\prod b_ t$. This proves (2).

Proof of (3). Choose $t \in T$. There exists an integer $n \geq 0$ and a chain $a_ t = a_{t, 0}, a_{t, 1}, \ldots , a_{t, n} = b_ t$ such that for every $1 \leq i \leq n$ either there is a homotopy from $a_{t, i - 1}$ to $a_{t, i}$ or there is a homotopy from $a_{t, i}$ to $a_{t, i - 1}$. If $n = 0$, then we pick another $t$. (We're done if $a_ t = b_ t$ for all $t \in T$.) So assume $n > 0$. By Example 14.26.3 there are is a homotopy from $b_{t'}$ to $b_{t'}$ for all $t' \in T \setminus \{ t\} $. Thus by (2) there is a homotopy from $a_{t, n - 1} \times \prod _{t'} b_{t'}$ to $b$ or there is a homotopy from $b$ to $a_{t, n - 1} \times \prod _{t'} b_{t'}$. In this way we can decrease $n$ by $1$. This proves (3).

Part (1) follows from part (3) and the definitions. $\square$

[1] In the literature, often the maps $h_{n + 1, i} \circ s_ i : U_ n \to V_{n + 1}$ are used instead of the maps $h_{n, i}$. Of course the relations these maps satisfy are different from the ones in Lemma 14.26.2.

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