*Natural: endogenous: produced by factors inside the system*

Throughout this page, we shall often denote by $\mathbf{X}_0$ the set of objects of a category $\mathbf{X}$ and by $F_0:\mathbf{X}_0\to \mathbf{Y}_0$ the map induced by a functor $F:\mathbf{X}\to \mathbf{Y}$.

If $\mathbf{X}$ and $\mathbf{Y}$ are categories, we shall say that a functor $F:\mathbf{X}\to \mathbf{Y}$ is an **isofibration** if for every object $A\in \mathbf{X}$ and every isomorphism $v\in \mathbf{Y}$ with source $F A$, there exists an isomorphism $u\in \mathbf{X}$ with source $A$ such that $F(u)=v$,

When the isomorphism $u$ is uniquely determined by a pair $(A,v)$, we shall say that the isofibration has the *unique lifting property*.

The notion of isofibration is self dual: a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff the opposite functor $F^o:\mathbf{X}^o\to \mathbf{Y}^o$ is an isofibration (exercise ). Hence a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff for every object $A\in \mathbf{X}$ and every isomorphism $v\in \mathbf{Y}$ with *target* $F A$, there exists an isomorphism $u\in \mathbf{X}$ with *target* $A$ such that $F(u)=v$,

Let $J$ be the groupoid generated by one isomorphism $0\simeq 1$. Then a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff it has the right lifting property with respect to the inclusion $\{0\}\subset J$ (resp. $\{1\}\subset J$). (exercise ).

We shall say that a functor $F:\mathbf{X}\to\mathbf{Y}$ is *monic* (resp. *surjective*, *bijective*) *on objects* if the map $\mathbf{X}_0\to \mathbf{Y}_0$ induced by $F$ is injective (resp. surjective, bijective).

Let us denote by $\mathbf{Cat}$ the category of small categories and by $\mathbf{Cat'}$ the category of locally small categories. The category $\mathbf{Cat'}$ cannot carry a model structure since it is not finitely cocomplete. However,

Let us denote by $\mathcal{C}'$ the class of functors monic on objects in the category $\mathbf{Cat'}$, by $\mathcal{W}'$ the class of equivalences, and by $\mathcal{F}'$ the class of isofibrations. Then the pairs $(\mathcal{C}'\,\cap\,\mathcal{W}',\mathcal{F}')$ and $(\mathcal{C}',\mathcal{F}'\,\cap\,\mathcal{W}')$ are weak factorisation systems in $\mathbf{Cat'}$. If

$\mathcal{C}=\mathcal{C}'\,\cap \, \mathbf{Cat},\quad
\quad \mathcal{W}=\mathcal{W}'\,\cap \, \mathbf{Cat}\quad
\mathrm{and} \quad \mathcal{F}=\mathcal{F}'\,\cap \, \mathbf{Cat}$

then the triple $(\mathcal{C},\mathcal{W},\mathcal{F})$ is a model structure on the category $\mathbf{Cat}$. The model structure is cartesian closed and proper. We shall say that it is the **natural model structure** on $\mathbf{Cat}$.

The proof will be given after Proposition .

The class $\mathcal{W}'$ has the three-for-two property and it is closed under retracts.

Let us write $F\simeq G$ to indicate that the functors $F,G:\mathbf{X}\to \mathbf{Y}$ are isomorphic in the category $[\mathbf{X},\mathbf{Y}]$. The relation $F\simeq G$ is compatible with composition on both sides: we have

$F\simeq G \quad \Rightarrow \quad L F\simeq L G \quad \mathrm{and} \quad
F R \simeq K R$

for every functor $L:\mathbf{Y}\to \mathbf{Y'}$ and every functor $R:\mathbf{X'}\to \mathbf{X}$. We can thus construct a quotient category $\mathrm{Ho}(\mathbf{Cat'})$ by putting

$\mathrm{Ho}(\mathbf{Cat'})(\mathbf{X},\mathbf{Y})=\mathbf{Cat'}(\mathbf{X},\mathbf{Y})/\simeq$

for locally small categories $\mathbf{X}$ and $\mathbf{Y}$. The canonical functor $Ho:\mathbf{Cat'}\to \mathrm{Ho}(\mathbf{Cat'})$ takes a functor $F:\mathbf{X}\to \mathbf{Y}$ to its isomorphism class in the category of functors $\mathbf{X}\to \mathbf{Y}$. The morphism $Ho(F):\mathbf{X}\to \mathbf{Y}$ is invertible iff the functor $F$ is an equivalence of categories. It follows that the class $\mathcal{W}'$ has the three-for-two property, since this is true of the class of isomorphisms in any category. Similarly, the class $\mathcal{W}'$ is closed under retracts, since this is true of the class of isomorphisms in any category.

For any functor $F:\mathbf{X}\to \mathbf{Y}$, any map $G_0:\mathbf{X}_0\to \mathbf{Y}_0$ and any family of isomorphisms $\theta=(\theta_A:F_0 A\to G_0 A| A\in \mathbf{X}_0)$ there exists a unique functor $G:\mathbf{X}\to \mathbf{Y}$ extending $G_0$ for which the family $\theta$ becomes a natural isomorphism $F\to G$.

The functor $G$ takes a map $f:A\to B$ in $\mathbf{X}$ to the unique map $G(f):G_0 A \to G_0 B$ in $\mathbf{Y}$ for which the following diagram commutes,

We shall say that the functor $G:\mathbf{X}\to \mathbf{Y}$ is obtained by **transporting** the functor $F:\mathbf{X}\to \mathbf{Y}$ along the family of isomorphisms $(\theta_A:F_0 A\to G_0 A| A\in \mathbf{X}_0)$. The possibility of transporting a functor $F:\mathbf{X}\to \mathbf{Y}$ along a family of isomorphisms $\theta$ actually means that the restriction functor $[\mathbf{X},\mathbf{Y}]\to [\mathbf{X}_0,\mathbf{Y}]$ is an isofibration with unique lifting, where $\mathbf{X}_0$ denote the discrete category whose objects are the elements of $\mathbf{X}_0$.

A functor is monic iff it is monic on objects and faithful. We shall say that a functor is *surjective* if it is surjective on objects and full.

A surjective equivalence $P:\mathbf{X}\to \mathbf{Y}$ is a split epimorphism; more precisely, every section $S_0:\mathbf{Y}_0 \to \mathbf{X}_0$ of the map $P_0:\mathbf{X}_0\to \mathbf{Y}_0$ can be extended uniquely as a section $S:\mathbf{Y}\to \mathbf{X}$ of the functor $P$. There then is a unique isomorphism $\theta:Id_{\mathbf{X}}\to S F$ such that $P\circ\theta=id_P$. Dually, a monic equivalence $U:\mathbf{A}\to \mathbf{B}$ is a split monomorphism. If $R:\mathbf{B}\to \mathbf{A}$ is a retraction, then there is a unique isomorphism $\theta:Id_{\mathbf{B}}\to U R$ such that $\theta \circ U=id_U$.

Let us prove the first statement. Let $P:\mathbf{X}\to \mathbf{Y}$ be an equivalence surjective on objects. The map $P_0:\mathbf{X}_0\to \mathbf{Y}_0$ has a section, since it is surjective. Let $S_0:\mathbf{Y}_0 \to \mathbf{X}_0$ be a section. If $f:A\to B$ is a morphism in $\mathbf{Y}$, then there is a unique morphism $S(f):S_0 A \to S_0 B$ in $\mathbf{Y}$ such that $P S(f)=f$, since the map $\mathbf{X}(S_0 A,S_0 B)\to \mathbf{Y}(A,B)$ induced by $P$ is bijective. This defines a functor $S:\mathbf{Y}\to \mathbf{X}$. It is easy to see that we have $P S=Id_{\mathbf{Y}}$. The functor $S$ is an equivalence of categories, since $P$ is an equivalence. The existence and uniqueness of an isomorphism $\theta:Id_{\mathbf{X}}\to S P$ such that $P\circ\theta=id_P$ follows. Let us prove the second statement. The functor $U$ is essentially surjective, since it is fully faithful. Thus, for each object $B \in \mathbf{B}_0$ we can choose an object $R_0B \in \mathbf{A}_0$ together with an isomorphism $\theta_B:B\to U_0R_0B$, with the proviso that $\theta_B=1_B$ when $B \in U(\mathbf{A}_0)$. The proviso implies that $R_0U_0A =A$ for every object $A\in \mathbf{A}_0$, since $U_0$ is monic. There is then a unique functor $R:\mathbf{B}\to \mathbf{A}$ extending $R_0$ for which the family $(\theta_B)$ becomes a natural isomorphism $id_{\mathbf{B}}\to U R$. By construction, the functor $R$ takes a morphism $g:B\to B'$ in $\mathbf{B}$ to the unique morphism $R(g):R_0 B\to R_0 B'$ such that the following square commutes,

If $g=U(f)$, where $f:A\to A'$ is a morphism in $\mathbf{A}$, then we have $R(g)=f$, since the morphisms $\theta_{B}$ are units when $B\in U( \mathbf{A})$ and the following square commutes,

This shows that $R U=id_{\mathbf{A}}$. The functor $R$ is an equivalence, since $U$ is an equivalence. The existence and uniqueness of an isomorphism $\theta:Id_{\mathbf{B}}\to U R$ such that $\theta \circ U=id_U$ follows.

The class of isofibrations is closed under composition, retracts and base changes.

Let $J$ be the groupoid generated by one isomorphism $0\simeq 1$. Then a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff it has the right lifting property with respect to the inclusion $i_0:\{0\}\subset J$. Thus, the class of isofibrations is of the form $\{i_0\}^\pitchfork$. The result then follows from the proposition here.

An equivalence is an isofibration iff it is surjective on objects. The class of equivalences surjective on objects is closed under composition, retracts and base changes.

Let us prove the first statement. Let $F:\mathbf{X}\to \mathbf{Y}$ be an equivalence which is an isofibration. Then for every object $B\in \mathbf{Y}$, there exists an object $A\in \mathbf{X}$ together with an isomorphism $v: F A\to B$, since an equivalence is essentially surjective. There is then an isomorphism $u:A\to A'$ such that $F(u)=v$, since $F$ is an isofibration. We then have $F A'=B$, and this shows that $F$ is surjective on objects. Conversely, let us show that an equivalence $F:\mathbf{X}\to \mathbf{Y}$ surjective on objects is an isofibration. If $A$ is an object of $\mathbf{X}$ and $v:F A \to B$ is an isomorphism in $Y$, then there exists an object $A'\in X$ such that $F A'=B$, since $F$ is surjective on object. The map $\mathbf{X}(A,A')\to \mathbf{Y}(F A,F A')$ induced by $F$ is bijective, since $F$ is an equivalence. Hence there exists a morphism $u: A\to A'$ such that $F(u)=v$. The morphism $u$ is invertible, since $v$ is invertible and $F$ is an equivalence.

This shows that $F$ is an isofibration. The first statement of the proposition is proved. Let us prove the second statement. Observe that a functor is an equivalence surjective on objects iff it is fully faithful and surjective on objects. Hence the class of equivalences surjective on objects is the intersection of two classes: the class $\mathcal{M}$ of fully faithful functors and the class $\mathcal{N}$ of functors surjective on objects. The class $\mathcal{M}$ is the right class of the Gabriel factorisation system by the Example here. It then follows from the propositions here and here that the class $\mathcal{M}$ is closed under composition, retracts and base changes. The class of surjections in the category of sets has the same closure properties, hence also the class $\mathcal{N}$.

Suppose that we have a commutative square of categories and functors

in which the functor $U$ is monic on objects and the functor $F$ is an isofibration. If $U$ or $F$ is an equivalence, then the square has a diagonal filler.

For this it suffices to show that the left hand square of the following diagram has a diagonal filler,

(1)

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The projection $pr_1$ is an isofibration by Lemma , since it is a base change of the functor $P$. Moreover, $pr_1$ is an equivalence when $P$ is an equivalence by Lemma{trivialfibrationclosure}. This shows that problem can be reduced to the case of a square

(2)

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Let us first consider the case where the functor $U$ is an equivalence. It follows from Lemma that the functor $U$ admits a retraction $R:\mathbf{B}\to \mathbf{A}$ together with a natural isomorphism $\theta:Id_{\mathbf{B}}\to U R$ such that $\theta\circ U =id_U$. This last condition means that we have $\theta_{U A}=1_{U A}$ for every object $A\in \mathbf{A})$. We have $P F R B=U R B$ for every object $B\in \mathbf{B}$, since $P F=U$. We thus have a diagram,

There exists an isomorphism $\alpha_B$ with source $F R B$ such that $P(\alpha_B)=\theta_B$, since the functor $P$ is an isofibration by assumption,

This defines a map $D_0:\mathbf{B}_0\to \mathbf{X}_0$, where $D_0 B$ is the target of $\alpha_B$. The isomorphism $\alpha_B$ can be taken to be a unit when $B=U(A)\in U(\mathbf{A}_0)$, since $\theta_B$ is a unit in this case. We then have $D_0 U_0(A)=F_0(A)$ for every $A\in \mathbf{A}_0$. If we transport the functor $F R : \mathbf{B}\to \mathbf{X}$ by Lemma along the the family of isomorphisms $(\alpha_B)$ we obtain a functor $D:\mathbf{B}\to \mathbf{X}$ equipped with an isomorphism $\alpha:F R\to D$. By construction, the functor $D$ takes a morphism $u:B\to B'$ in $\mathbf{B}$ to the unique morphism $D(u):D B\to D B'$ in $\mathbf{X}$ such that the following square commutes,

It is then a routine matter to verify that we have $D U=F$ and $P D =Id_{\mathbf{B}}$. We have proved that the square (2) has a diagonal filler in the case where $U$ is an equivalence. Let us now consider the case where $P$ is an equivalence. The functor $P$ is surjective on objects in this case by Lemma . Hence the following square in the category of sets

has a diagonal filler $D_0:\mathbf{B}_0\to \mathbf{X}_0$, since $U_0$ is monic on objects and the pair $(Inj,Surj)$ is a weak factorisation system in the category of sets. It then follows from Lemma that the functor $P$ admits a unique section $D:\mathbf{B}\to \mathbf{X}$ which extends the map $D_0$, since $P$ is a surjective equivalence. It is then a routine matter to verify that we have $D U=F$. We have proved that the square (2) has a diagonal filler in the case where $P$ is an equivalence.

Let $J$ be the groupoid generated by one isomorphism $0\simeq 1$. We shall denote the inclusion $\{0\}\subset J$ as a map $d_1:1\to J$ and the inclusion $\{1\}\subset J$ as a map $d_0:1\to J$. This is in accordance with the standard notation for the maps $d_0,d_1:[0]\to [1]$ in the category $\Delta$, since $d_0$ takes the value $1$ and $d_1$ takes the value $0$.

The path object? of a category $\mathbf{X}$ is defined to be the category $\mathbf{X}^J=[J,\mathbf{X}]$. An *object* of this category is an isomorphism $a:A_0\to A_1$ in the category $\mathbf{X}$, and a *morphism* $u:a\to b$ between $a$ and $b:B_0\to B_1$ is a pair $(u_0,u_1)$ of maps in a commutative square

(3)

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The functor $\partial_1=[d_1,\mathbf{X}]:[J,\mathbf{X}]\to \mathbf{X}$ is the *source functor* which which takes an isomorphism $a:A_0\to A_1$ to its source $A_0$, and $\partial_0=[d_0,\mathbf{X}]$ is the *target functor* which which takes an isomorphism $a:A_0\to A_1$ to its target $A_1$. If $s$ denotes the functor $J\to 1$, then the functor $\sigma=[s,\mathbf{X} ]:\mathbf{X}\to [J,\mathbf{X}]$ is the *unit functor* which takes an object $A\in \mathbf{X}$ to the unit isomorphism $1_A:A\to A$. The relation $s d_1 =id_1=s d_0$ implies that we have $\partial_1 \sigma =id_{\mathbf{X}}=\partial_0 \sigma$. The functors $\partial_1,\partial_0$ and $\sigma$ are equivalences of categories, since the functors $d_1,d_0$ and $s$ are equivalences.

The functor

$(\partial_1,\partial_0):\mathbf{X}^J\to \mathbf{X}\,\times\, \mathbf{X}$

is an isofibration and the functor $\sigma:\mathbf{X}\to \mathbf{X}^J$ is an equivalence of categories. Moreover, the functors $\partial_1$ and $\partial_0$ are equivalences surjective on objects.

Let us show that the functor $(\partial_1,\partial_0)$ is an isofibration. Let $a:A_0\to A_1$ be an object of $\mathbf{X}^J$ and let $(u_0,u_1):(A_0,A_1)\to (B_0,B_1)$ be an isomorphism in $\mathbf{X}\times \mathbf{X}$. There is then a unique isomorphism $b:B_0\to B_1$ such that the square (3) commutes. The pair $u=(u_0,u_1)$ defines an isomorphism $a\to b$ in the category $\mathbf{X}^J$, and we have $(\partial_1,\partial_0)(u)=(u_0,u_1)$. This proves that $(\partial_1,\partial_0)$ is an isofibration. We saw above that the functor $\partial_1,\partial_0$ and $\sigma$ are equivalences of categories. The functor $\partial_1$ is surjective on objects, since $\partial_1\sigma=id_{\mathbf{X}}$. Similarly, the functor $\partial_0$ is surjective on objects.

The mapping path object? of a functor $F:\mathbf{X}\to \mathbf{Y}$ is the category $\mathbf{P}(F)$ defined by the following pullback square

(4)

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There is a (unique) functor $i_{\mathbf{X}}:\mathbf{X}\to \mathbf{P}(F)$ such that $P i_{\mathbf{X}}=\sigma F$ and $P_{\mathbf{X}} i_{\mathbf{X}}=id_{\mathbf{X}}$ since square (4) is cartesian and we have $\partial_1\sigma F=id_{\mathbf{Y}} F =F id_{\mathbf{X}}$. Let us put $P_{\mathbf{Y}}=\partial_0 P$. Then we have

$F=P_{\mathbf{Y}} i_{\mathbf{X}}:\mathbf{X}\to \mathbf{P}(F)\to \mathbf{Y}$

since $P_{\mathbf{Y}} i_{\mathbf{X}}=\partial_0 P i_{\mathbf{X}}=\partial_0 \sigma F=id_{\mathbf{Y}} F =F.$ This is the mapping path factorisation? of the functor $F$. Let us describe the category $\mathbf{P}(F)$ explicitly. Let us first show that we have a pullback square,

(5)

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The square commutes, since $P_{\mathbf{Y}}=\partial_0 P$. To see that it is cartesian, consider the diagram

The right hand square of this diagram is trivially cartesian. The composite square is cartesian by definition of $\mathbf{P}(F)$. Hence the left hand square is also cartesian by the lemma here. This shows that the square (5) is cartesian. We now use it for describing the objects and morphisms of the category $\mathbf{P}(F)$. By construction, an object of $\mathbf{P}(F)$ is a triple $(y,A,B)$, where $A$ is an object of $\mathbf{X}$, $B$ is an object of $\mathbf{Y}$ and $y$ is an isomorphism $F(A)\to B$. The object can be pictured as a leg with the upper part in $\mathbf{X}$ and with its foot in $\mathbf{Y}$:

We have $P(y,A,B)=y$, $P_{\mathbf{X}}(y,A,B)=A$ and $P_{\mathbf{Y}}(y,A,B)=B$. A *morphism* $(y,A,B)\to (y',A',B')$ in the category $\mathbf{P}(F)$ is a pair of maps $u:A\to A'$ and $v:B\to B'$ such that the square foot of the following diagram commutes,

The functor $i_{\mathbf{X}}:\mathbf{X}\to \mathbf{P}(F)$ takes an object $A\in \mathbf{X}$ to a leg with a very short foot,

The mapping path category $\mathbf{P}(F)$ can be constructed as the pseudo-pullback? of the functor $F: \mathbf{X}\to \mathbf{Y}$ with the identity functor $\mathbf{Y}\to \mathbf{Y}$.

The functor $P_{\mathbf{Y}}$ in the mapping path factorisation

$F=P_{\mathbf{Y}} i_{\mathbf{X}}:\mathbf{X}\to \mathbf{P}(F)\to \mathbf{Y}$

is an isofibration and the functor $i_{\mathbf{X}}$ is an equivalence monic on objects.

Let us show that the functor $P_{\mathbf{Y}}$ is an isofibration. We first give a formal proof by using the general argumentation of Quillen. The functor $(P_{\mathbf{X}},P_{\mathbf{Y}})$ is a base change of the functor $(\partial_1,\partial_0)$, since the square (5) is cartesian. Hence the functor $(P_{\mathbf{X}},P_{\mathbf{Y}})$ is an isofibration by Proposition , since the functor $(\partial_1,\partial_0)$ is an isofibration by Proposition . The projection $pr_2:\mathbf{X}\,\times\, \mathbf{Y}\to \mathbf{Y}$ is a base change of the functor $\mathbf{X}\to 1$. It is thus an isofibration, since the functor $\mathbf{X}\to 1$ is (trivially) an isofibration. It follows that the composite $P_{\mathbf{Y}}=pr_2(P_{\mathbf{X}},P_{\mathbf{Y}})$ is an isofibration, since the class of isofibrations is closed under composition by Proposition . Let us now give the bare foot proof. For every object $(u,A,B)\in \mathbf{P}(F)$ and every isomorphism $y:B\to B'$ the pair $(1_A,y)$ is an isomorphism $(u,A,B)\to (y u,A, B')$ and we have $P_{\mathbf{Y}}(1_A,y)=y$,

This proves that $P_{\mathbf{Y}}$ is an isofibration. It remains to prove that the functor $i_{\mathbf{X}}$ is an equivalence monic on objects. It is certainly monic on objects, since we have $P_{\mathbf{X}}i_{\mathbf{X}}=id_{\mathbf{X}}$. Let us show that it is an equivalence. Again, we shall give two proofs, the first by using the general argumentation of Quillen. It suffices to prove that the functor $P_{\mathbf{X}}$ is an equivalence by three-for-two, since we have $P_{\mathbf{X}}i_{\mathbf{X}}=id_{\mathbf{X}}$. But the functor $P_{\mathbf{X}}$ is a base change of the functor $\partial_1:\mathbf{Y}^J\to \mathbf{Y}$, since the square (4) is cartesian. The functor $\partial_1$ is an equivalence surjective on objects by Proposition . It follows that the functor $P_{\mathbf{X}}$ is an equivalence by Proposition . Let us now gives the bare foot proof that the functor $i_{\mathbf{X}}$ is an equivalence. For this it suffices to exibit a natural isomorphism $\alpha: i_{\mathbf{X}} P_{\mathbf{X}} \simeq id$ since we already have $P_{\mathbf{X}}i_{\mathbf{X}}=id_{\mathbf{X}}$. But $i_{\mathbf{X}} P_{\mathbf{X}}(u,A,B)=(1_{F A},A,F A)$ and the pair $(1_A,u)$ defines a *natural* isomorphism $(1_{F A}, A, F A)\to (u,A,B)$,

The class of functors monic on objects is closed under composition, retracts and cobase changes.

The class of injective maps in the category of sets is closed under composition, retracts and cobase changes. Hence also the class of functors monic on objects.

The category $\mathbf{Cat'}$ admits a weak factorisation system $(\mathcal{L}, \mathcal{R})$ in which $\mathcal{L}$ is the class of equivalences monic on objects and $\mathcal{R}$ is the class of isofibrations.

We shall use the characterisation of weak factorisation systems here. The mapping path category $\matbf{P}(F)$ of a functor $F:\mathbf{X} \to \mathbf{Y}$ is locally small if the category $\mathbf{X}$ is locally small, since the functor $i_{\mathbf{X}}:\mathbf{X} \to \mathbf{P}(F)$ is an equivalence by Proposition . It then follows from this proposition that every functor $F:\mathbf{X}\to \mathbf{Y}$ in $\mathbf{Cat'}$ admits a $(\mathcal{L}, \mathcal{R})$-factorisation. By Lemma we have $\mathcal{L}\,\pitchfork\, \mathcal{R}$. It remains to verify that the classes $\mathcal{L}$ and $\mathcal{R}$ are closed under retracts. This is true for the class $\mathcal{R}$ by Proposition . And this is true for the class of functors monic on objects by Lemma . Moreover, a retract of an equivalence is an equivalence by Lemma .

The class of equivalences monic on objects is closed under composition, retracts and cobase changes.

The class of equivalences monic on objects is the left class of a weak factorisation system by Proposition . It is thus closed under composition, retracts and cobase changes.

The cylinder? of a category $\mathbf{A}$ is defined to be the category $J \mathbf{A}= J\times \mathbf{A}$. We shall denote the inclusion $\{0\}\times \mathbf{A} \subseteq J\times \mathbf{A}$ by $d_1:\mathbf{A}\to J A$, the inclusion $\{1\}\times \mathbf{A} \subseteq J\times \mathbf{A}$ by $d_0:\mathbf{A} \to J \mathbf{A}$, and the projection $pr_2:J\times \mathbf{A}\to \mathbf{A}$ by $s:J \mathbf{A}\to \mathbf{A}$. Notice that $s d_1=id_{\mathbf{A}}=s d_0$.

The functor $(d_1,d_0): \mathbf{A}\sqcup \mathbf{A}\to J \mathbf{A}$ is monic on objects and the projection $s:J \mathbf{A}\to \mathbf{A}$ is an equivalence. Moreover, the functors $d_1$ and $d_0$ are equivalences monic on objects.

I spare you the proof.

The mapping cylinder? of a functor $F:\mathbf{A}\to \mathbf{A}$ is the category $\mathbf{C}(F)$ defined by the following pushout square:

(6)

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There is a (unique) functor $Q_{\mathbf{B}}:\mathbf{C}(F)\to \mathbf{B}$ such that $Q_{\mathbf{B}}Q=F s$ and $Q_{\mathbf{B}} i_{\mathbf{B}}=id_{\mathbf{B}}$, since square (6) is cocartesian and we have $F s i_1=F id_{\mathbf{A}}=id_{\mathbf{B}}F$. Let us put $i_{\mathbf{A}}:=Q d_1$. Then we have

$F=Q_{\mathbf{B}} i_{\mathbf{A}}:\mathbf{A}\to \mathbf{C}(F)\to \mathbf{B},$

since $Q_{\mathbf{B}} i_{\mathbf{A}}=Q_{\mathbf{B}} Q d_1=F s d_1=F id_{\mathbf{A}}=F$. This is the mapping cylinder factorisation? of the functor $F$. Let us show that we have a pushout square,

(7)

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The square commutes since $i_{\mathbf{A}}=Q d_1$. The square is a pushout, since both the top square and the composite square of the following diagram are cocartesian,

(8)

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In the mapping cylinder factorisation,

$F=Q_{\mathbf{B}} i_{\mathbf{A}}:\mathbf{A}\to \mathbf{C}(F)\to \mathbf{B},$

the functor $i_{\mathbf{A}}$ is monic on objects and the functor $Q_{\mathbf{B}}$ is an equivalence surjective on objects.

The functor $(i_{\mathbf{A}},i_{\mathbf{B}})$ is a cobase change of the functor $(d_1,d_0)$, since the square (7) is cocartesian. Hence the functor $(i_A,i_B)$ is monic on objects by Proposition , since the functor $(d_1,d_0)$ is monic on objects by Proposition . It follows that the functor $i_{\mathbf{A}}=(i_{\mathbf{A}},i_{\mathbf{B}})in_1$ is monic on objects since the functor $in_1:A\to A\sqcup B$ is monic on objects. Let us now show that the functor $Q_{\mathbf{B}}$ is an equivalence surjective on objects. The functor is obviously surjective on objects, since we have $Q_{\mathbf{B}}i_{\mathbf{B}}=id_{\mathbf{B}}$. In order to show that it is an equivalence, it suffices to show that the functor $i_{\mathbf{B}}$ is an equivalence by three-for-two. But $i_{\mathbf{B}}$ is a cobase change of the functor $\partial_0$, since the square (6) is cocartesian. It follows that the functor $i_{\mathbf{B}}$ is an equivalence by the Proposition , since $\partial_0$ is an equivalence monic on objects by Proposition .

The category $\mathbf{Cat'}$ admits a weak factorisation system $(\mathcal{L}, \mathcal{R})$ in which $\mathcal{L}$ is the class functors monic on objects and $\mathcal{R}$ is the class of acyclic isofibration.

We shall use the characterisation of weak factorisation systems here. The mapping cylinder $\mathbf{C}(F)$ of a functor $F:\mathbf{A} \to \mathbf{B}$ is locally small if the category $\mathbf{B}$ is locally small, since the functor $Q_{\mathbf{B}}:\mathbf{C}(F)\to\mathbf{B}$ is an equivalence by Proposition . The same proposition shows that every functor $F:\mathbf{A} \to \mathbf{B}$ admits a $(\mathcal{L}, \mathcal{R})$-factorisation. It follows from Lemma that we have $\mathcal{L}\,\pitchfork\, \mathcal{R}$. The class $\mathcal{L}$ is closed under codomain retracts by Corollary . Also the class $\mathcal{R}$ by Lemma .

The class $\mathcal{W}$ of equivalences in $\mathbf{Cat}$ has the three-for-two property by Lemma . The pair $(\mathcal{C}\,\cap\, \mathcal{W}, \mathcal{F})$ is a weak factorisation system by Proposition , and the pair $(\mathcal{C}, \mathcal{F}\,\cap\, \mathcal{W})$ is a weak factorisation system by Proposition . This completes the proof that the triple $(\mathcal{C}\,\cap\, \mathcal{W}, \mathcal{F})$ is a model structure on the category $\mathbf{Cat}$. Every object $\mathbf{X}$ of this model structure is fibrant and cofibrant, since the map $\mathbf{X}\to 1$ is an isofibration and the map $\emptyset \to \mathbf{X}$ is monic on objects. Hence the model structure is proper by a proposition [here]. Let us show that the model structure is cartesian closed. If $S:\mathbf{A}\to \mathbf{B}$ and $T:\mathbf{U}\to \mathbf{V}$, are two functors, consider the functor

$S\times'T:(\mathbf{A}\times \mathbf{V})
\sqcup_{\mathbf{A}\times \mathbf{U}}(\mathbf{B}\times \mathbf{U})\to \mathbf{B}\times \mathbf{V}$

obtained from the commutative square

(9)

```
```

Let us show that the functor $S\times'T$ is a cofibration if $S$ and $T$ are cofibrations, and that it is a an equivalence if in addition $S$ or $T$ is an equivalence. We have $(S\times'T)_0=S_0\times'T_0$, since the functor $Ob:\mathbf{Cat}\to \mathbf{Set}$ preserves cartesian products and pushouts (it is actually continuous and cocontinuous). But the map $S_0\times'T_0$ is monic, if $S_0$ and $T_0$ are monic by a result [here]. This shows that $S\times'T$ is a cofibration if $S$ and $T$ are cofibrations. If in addition $T$ is an equivalence, then so are the vertical maps in the square, since the class of equivalences is closed under products. Consider the following diagram with a pushout square on the left hand side,

The functor $in_2$ is an acyclic cofibration by cobase change, since the functor $\mathbf{A}\times T$ is. Hence the functor $S\times' T$ is acyclic by three-for-two, since the functors $in_2$ and $\mathbf{B}\times T$ are acyclic.

If $U:\mathbf{A}\to \mathbf{B}$ and $F:\mathbf{X}\to \mathbf{Y}$, are two functors, consider the functor

$[ U,F]':[\mathbf{B},\mathbf{X}]
\to [\mathbf{B},\mathbf{Y}]
\times_{[\mathbf{A},\mathbf{Y}]}[\mathbf{A},\mathbf{X}]$

obtained from the commutative square

Recall that the category $[\mathbf{A},\mathbf{X}]$ is locally small (resp. small) if $\mathbf{A}$ is small and $\mathbf{X}$ is locally small (resp. small).

The functor $\mathbf{[} U,F\mathbf{]}'$ is an isofibration, if the functor $F$ is an isofibration and the functor $U$ is monic on objects. Moreover, the functor $\mathbf{[} U,F\mathbf{]}'$ is an equivalence, if in addition one of the fonctors $U$ or $F$ is an equivalence.

Recall that a functor $F:\mathbf{A}\to \mathbf{B}$ between small categories is said to be a **Morita equivalence** if the inverse image functor

$F^*:[\mathbf{B}^o,\mathbf{Set}]\to [\mathbf{A}^o,\mathbf{Set}]$

is an equivalence of categories. A functor $F:\mathbf{A}\to \mathbf{B}$ is a Morita equivalence iff it is fully faithful and every object $B\in \mathbf{B}$ is a retract of an object in the image of $F$.

Recall an idempotent $e:B\to B$ in a category $\mathbf{C}$ is said to **split** if there exists a pair of morphisms $s:A\to B$ and $r:B\to A$ such that $e=s r$ and $rs=1_A$. We shall say that $\mathbf{C}$ is **Karoubi complete** if every idempotent in $\mathbf{C}$ splits.

Let $Spl$ be the category freely generated by two arrows $s:0\to 1$ and $r:1\to 0$ such that $r s=1_0$. And let $Idem$ be the category freely generated by one idempotent $e:1\to 1$. Then the functor $idem \to Split$ which takes $e$ to $s r$ is fully faithful.

We shall say that a functor is an **idfibration** if it has the right liffting property with respect to the inclusion $Idem \to Spl$. A category $\mathbf{C}$ is Karoubi complete iff the functor $\mathbf{C}\to \mathbf{1}$ is an idfibration.

An idfibration is an isofibration. It is easy to see that an isofibration $F:\mathbf{X}\to \mathbf{Y}$ is an idfibration iff the functor $F$ reflects split idempotents, that is, if the implication

$F(e)\quad \mathrm{splits} \quad \Rightarrow
\quad e \quad \mathrm{splits}$

is true for every idempotent $e\in \mathbf{X}$.

The category of small categories $\mathbf{Cat}$ admits a model structure $(\mathcal{C},\mathcal{W},\mathcal{F})$ in which $\mathcal{C}$ is the class of functors monic on objects, $\mathcal{W}$ is the class of Morita equivalences and $\mathcal{F}$ is the class of idfibrations. The model structure is cartesian closed and left proper. We shall say that it is the **Karoubian model structure** on the category $\mathbf{Cat}$.

Show that the notion of isofibration is self-dual: a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff the opposite functor $F^o:\mathbf{X}^o\to \mathbf{Y}^o$ is an isofibration.

Show that a functor $F:\mathbf{X}\to \mathbf{Y}$ is an isofibration iff it has the right lifting property with respect to the inclusion $\{0\}\subset J$.

Revised on November 20, 2020 at 21:49:59
by
Dmitri Pavlov