Proof

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

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

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.

Proof

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,

Proof

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.

Proof

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.

Proof

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 $\mathcal{N}$.

Proof

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

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

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.

Proof

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.

Proof

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 $(1_{F A}, A, F A)\to (u,A,B)$,

Proof

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.

Proof

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 .

Proof

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.

Proof

I spare you the proof.

Proof

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 .

Proof

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 .

Proof of Theorem 1.

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

obtained from the commutative square

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.