for ∞-groupoids

Contents

Idea

The extra structure of a model category over a category with weak equivalences induces concrete constructions for expressing homotopy between morphisms. These lead in particular to an explicit construction of the homotopy category of a model category.

Definition

Definition

Let $\mathcal{C}$ be a model category and $X \in \mathcal{C}$ an object.

• A path object $Path(X)$ for $X$ is a factorization of the diagonal $\nabla_X \colon X \to X \times X$ as
$\nabla_X \;\colon\; X \underoverset{\in W}{i}{\longrightarrow} Path(X) \overset{(p_0,p_1)}{\longrightarrow} X \times X \,.$

where $X\to Path(X)$ is a weak equivalence. This is called a good path object if in addition $Path(X) \to X \times X$ is a fibration.

• A cylinder object $Cyl(X)$ for $X$ is a factorization of the codiagonal (or “fold map”) $\Delta_X X \sqcup X \to X$ as
$\Delta_X \;\colon\; X \sqcup X \overset{(i_0,i_1)}{\longrightarrow} Cyl(X) \underoverset{p}{\in W}{\longrightarrow} X \,.$

where $Cyl(X) \to X$ is a weak equivalence. This is called a good cylinder object if in addition $X \sqcup X \to Cyl(X)$ is a cofibration.

Remark

By the factorization axioms every object in a model category has both a good path object and as well as a good cylinder object according to def. . But in some situations one is genuinely interested in using non-good such objects.

For instance in the classical model structure on topological spaces, the obvious object $X\times [0,1]$ is a cylinder object, but not a good cylinder unless $X$ itself is cofibrant (a cell complex in this case).

More generally, the path object $Path(X)$ of def. is analogous to the powering $\pitchfork(I,X)$ with an interval object and the cyclinder object $Cyl(X)$ is analogous to the tensoring with a cylinder object $I\odot X$. In fact, if $\mathcal{C}$ is a $V$-enriched model category and $X$ is fibrant/cofibrant, then these powers and copowers are in fact examples of (good) path and cylinder objects if the interval object is sufficiently good.

Definition

Let $f,g \colon X \longrightarrow Y$ be two parallel morphisms in a model category.

• A left homotopy $\eta \colon f \Rightarrow_L g$ is a morphism $\eta \colon Cyl(X) \longrightarrow Y$ from a cylinder object of $X$, def. , such that it makes this diagram commute:
$\array{ X &\longrightarrow& Cyl(X) &\longleftarrow& X \\ & {}_{\mathllap{f}}\searrow &\downarrow^{\mathrlap{\eta}}& \swarrow_{\mathrlap{g}} \\ && Y } \,.$
• A right homotopy $\eta \colon f \Rightarrow_R g$ is a morphism $\eta \colon X \to Path(Y)$ to some path object of $X$, def. , such that this diagram commutes:
$\array{ && X \\ & {}^{\mathllap{f}}\swarrow & \downarrow^{\mathrlap{\eta}} & \searrow^{\mathrlap{g}} \\ Y &\longleftarrow& Path(Y) &\longrightarrow& Y } \,.$

Properties

Basic lemmas

Lemma

If $X \sqcup X \overset{(i_0,i_1)}{\longrightarrow} Cyl(X) \underoverset{p}{\in W}{\longrightarrow} X$ is a good cylinder object for a cofibrant object $X$ def. , then both components $i_0, i_1 \colon X \to Cyl(X)$ are acyclic cofibrations.

Dually, if $X \underoverset{\in W}{i}{\longrightarrow} Path(X) \overset{(p_0,p_1)}{\longrightarrow} X \times X$

is a good path object for a fibrant object $X$, then both component $p_0,p_1 \colon Path(X)\to X$ are acyclic fibrations.

Proof

We discuss the first case, the second is formally dual. First observe that the two inclusions $X \to X \sqcup X$ are cofibrations, since they are the pushout of the cofibration $\emptyset \to X$. This implies that $i_0$ and $i_1$ are composites of two cofibrations

$i_0, i_1 \;\colon\; X \overset{\in Cof}{\longrightarrow} X\sqcup X \overset{\in Cof}{\longrightarrow}$

and hence are themselves cofibrations. That they are in addition weak equivalences follows from two-out-of-three applied to the identity

$id_X \;\colon\; X \overset{\in W}{\longrightarrow} Cyl(X) \overset{i_0}{\longrightarrow} X \,.$

implied by the fact that the cylinder by definition factors the codiagonal.

The following says that the choice of cylinder/path objects in def. is irrelevant as long it is “good”.

Lemma

For $\eta \colon f \Rightarrow_L g \colon X \to Y$ a left homotopy in some model category, def. , such that $Y$ is a fibrant object, then for $Cyl(X)$ any choice of good cylinder object for $X$, def. , there is a commuting diagram of the form

$\array{ X &\longrightarrow& Cyl(X) &\longleftarrow& X \\ & {}_{\mathllap{f}}\searrow &\downarrow^{\mathrlap{\tilde \eta}}& \swarrow_{\mathrlap{g}} \\ && Y } \,.$

Dually, for $\eta \colon f \Rightarrow_R g \colon X \to Y$ a right homotopy, def. , such that $X$ is cofibrant, then for $Path(X)$ any choice of good path object for $X$, def. , there is a commuting diagram of the form

$\array{ && X \\ & {}^{\mathllap{f}}\swarrow & \downarrow^{\mathrlap{\tilde \eta}} & \searrow^{\mathrlap{g}} \\ Y &\longleftarrow& Path(Y) &\longrightarrow& Y } \,.$
Proof

We discuss the first statement, the second is formally dual. Let $\eta \colon \hat X \longrightarrow Y$ be the given left homotopy with respect to a given cylinder object $\hat X$ of $X$. Factor it as

$\eta \;\colon\; \hat X \overset{\in Cof}{\longrightarrow} Z \overset{\in W \cap Fib}{\longrightarrow} Y \,.$

Then find liftings $\ell$ and $k$ in the following two commuting diagrams

$\array{ X \sqcup X &\overset{}{\longrightarrow}& \hat X &\longrightarrow& Z \\ \downarrow && & {}^{\mathllap{\ell}}\nearrow & \downarrow \\ Cyl(X) &\longrightarrow& &\longrightarrow& Y } \;\;\;\;\; \,, \;\;\;\;\; \array{ \hat X &\overset{\eta}{\longrightarrow}& Y \\ \downarrow &{}^{\mathllap{k}}\nearrow& \downarrow \\ Z &\longrightarrow& \ast } \,.$

Now the composite $\eta \coloneqq k \circ \ell$ is of the required kind,

$\array{ X \sqcup X &\overset{}{\longrightarrow}& \hat X &\longrightarrow& Z &\overset{k}{\longrightarrow}& Y \\ \downarrow &&& {}^{\mathllap{\ell}}\nearrow & \\ Cyl(X) &\longrightarrow& } \,.$
Lemma

Let $f,g \colon X \to Y$ be two parallel morphisms in a model category.

• Let $X$ be cofibrant. If there is a left homotopy $f \Rightarrow_L g$ then there is also a right homotopy $f \Rightarrow_R g$ (def. ) with respect to any chosen good path object.

• Let $Y$ be fibrant. If there is a right homotopy $f \Rightarrow_R g$ then there is also a left homotopy $f \Rightarrow_L g$ with respect to any chosen good cylinder object.

Proof

We discuss the first case, the second is formally dual. Let $\eta \colon Cyl(X) \longrightarrow Y$ be the given left homotopy. By lemma we may assume without restriction that $Cyl(X)$ is good in the sense of def. , for otherwise replace it by one that is. With this, lemma implies that we have a lift $h$ in the following commuting diagram

$\array{ X &\overset{i \circ f}{\longrightarrow}& Path(Y) \\ {}^{\mathllap{i_0}}_{\mathllap{\in W \cap Cof}}\downarrow &{}^{\mathllap{h}}\nearrow& \downarrow^{\mathrlap{p_0,p_1}}_{\mathrlap{\in Fib}} \\ Cyl(X) &\underset{(f \circ p,\eta)}{\longrightarrow}& Y \times Y } \,,$

where on the right we have the chosen path space object. Now the composite $\tilde \eta \coloneqq h \circ i_1$ is a right homotopy as required.

$\array{ && && Path(Y) \\ && &{}^{\mathllap{h}}\nearrow& \downarrow^{\mathrlap{p_0,p_1}}_{\mathrlap{\in Fib}} \\ X &\overset{i_1}{\longrightarrow}& Cyl(X) &\underset{(f \circ p,\eta)}{\longrightarrow}& Y \times Y } \,.$

Equivalence relation

Proposition

For $X$ a cofibrant object in a model category and $Y$ a fibrant object, then the relations of left homotopy $f \Rightarrow_L g$ and of right homotopy $f \Rightarrow_R g$ (def. ) on the hom set $Hom(X,Y)$ coincide and are both equivalence relations.

Proof

That both relations coincide under the (co-)fibrancy assumption follows directly from lemma .

To see that left homotopy with domain $X$ is a transitive relation first use lemma to obtain that every left homotopy is exhibited by a good cylinder object $Cyl(X)$ and then lemma to see that the cofiber coproduct $Cyl(X)\underset{X}{\sqcup} Cyl(X)$ in

$\array{ && && X \\ && && \downarrow^{\mathrlap{i_0}} \\ && X &\underoverset{\in W}{i_1}{\longrightarrow}& Cyl(X) \\ && {}^{\mathllap{i_0}}\downarrow &(po)& \downarrow \\ X &\underset{i_1}{\longrightarrow}& Cyl(X) &\underset{\in W}{\longrightarrow}& Cyl(X) \underset{X}{\sqcup} Cyl(X) \\ && &{}_{\mathllap{}}\searrow& & \searrow \\ && && \underset{\in W}{\longrightarrow} &\longrightarrow& X }$

is again a cylinder object, def. . The symmetry and reflexivity of the relation is obvious.

See the references at model category.