Simplicial

Proof. Clearly these are implied by the definition of a weak factorization system. Conversely, it will suffice by duality to show that if a morphism \(f\) has the left lifting property with respect to all morphisms in \(R\), it is in \(L\). But we can factor \(f = g\circ h\) with \(h \in L, g \in R\). But finding a lift in the diagram below shows \(f\) is a retract of \(h\).

□

Proof. The data of an adjunction is the same as giving the counit and unit maps that satisfy some identities. Since \(F,G\) are homotopical, they descent to functors on \(hC\) and \(hD\), and so do the unit and counits, and the diagrams still commute. □

Proof. Let \(f:X \to Y\) be our map and let \(A_f\) be the set of all commutative squares \(S\) of the form

where \(S_\alpha \in J\). Now consider the pushout diagram:

There is a canonical projection \(f_1:X_1\to Y\) given by the \(S_y\) and \(f\), and the construction of \(X_1\) is functorial. Moreover, \(X \to X_1\) is in \(\phantom {}^\oslash (J^{\oslash })\), and for any diagram in \(A\), there is a canonical lift on \(X_1\). Now we can inductively define \(f_\alpha , X_\alpha \) for any limit ordinal. Namely, \(X_{\alpha +1},f_{\alpha +1} = (X_{\alpha })_1,(f_{\alpha })_1\). For a limit ordinal, define it to be the limit. of the \(X_i,f_i\) before it. Then \(X \to X_{\kappa }\) is in \(\phantom {}^\oslash (J^{\oslash })\) as it is a transfinite composite of such things, and \(f_{\kappa }:X_{\kappa } \to Y\) is in \(J^{\oslash }\) since the codomains are \(\kappa \)-small.

This verifies factorization in a functorial way. The lifting property comes from Lemma 2.11, so this is a weak factorization system, and it is clearly the minimal one containing \(J\) on the left. We have only used the operations in the weak saturation of \(J\) to construct the factorization, so by Lemma 4.2, \(\phantom {}^\oslash (J^{\oslash })\) is the weak saturation of \(J\). □

Proof. Those conditions are necessary for weak equivalences to satisfy \(2\) out of \(3\). In the proof of sufficiency, we declare an arrow \(f\) to be a weak equivalence if it factors as \(h \circ f\), where \(h\) is a trivial fibration and \(f\) a triival cofibration. Because of \(3\) out of \(4\) and the fact that we have a factorization system, this is equivalent to \(f\) factoring as a composite of maps that are either trivial fibrations or trivial cofibrations.

First we observe that the trivial fibrations are the fibrations that are weak equivalences and dually for cofibrations. Indeed, a trivial fibration by assumption is a fibration, and the trivial factorization shows it is a weak equivalence. If \(f\) is a trivial fibration and a week equivalence, then factoring it as a trivial fibration and cofibration and lifting as in the diagram below shows that it is a retract of a trivial fibration.

It remains then to show weak equivalences satisfy \(2\) out of \(3\).

First we can observe that any composite of weak equivalences is a weak equivalence because it can be factored a composite of trivial fibrations and cofibrations. Now suppose \(h = g\circ f\) where \(f,h\) are weak equivalences. We can factor \(g\) into a fibration \(g'\) and trivial cofibration, which we can absorb into \(f\). Then factor \(f,h\) as \(f' \circ f'',h'\circ h''\), trivial cofibrations followed by a trivial fibration. Because of \(2\) out of \(3\), it suffices to show that \(g \circ f'\) is a trivial fibration. But this follows from the \(3\) out of \(4\) property on the square \((g \circ f') \circ f'' = h'\circ h''\). The other part of \(2\) out or \(3\) is dual. □

Proof. It suffices to check that \(\xrightarrow {\sim }\cap \rightarrowtail \subset \overset {\sim }{\rightarrowtail }\). But we can factor a weak equivalence as a triival cofibration and trivial fibration, and if it is also a cofibration lift the square of this factorization to see that it is a a retract of a trivial cofibration. □

Proof. The condition on fibrations and weak equivalences clearly determines the model structure if it exists. We can construct our factorizations via Proposition 4.8 since the generators of the factorization systems for \(D\) are \(F\) applied to the ones for \(C\) by the adjunction, and these are sent to \(\kappa \)-small objects. \(2\) out of \(3\) is clear so by Lemma 4.14 and the last assumption, we are done. □

Proof. Any solution of the lifting problem in \(C\) will actually be a solution in \(C/D\). □

Proof. This is a consequence of Lemma 4.18 and the fact that any factorzation will automatically becoma a relative factorization. □

Proof. Finite limits and colimits exist in \(C_{/D}\), and weak equivalences satisfy \(2\) out of \(3\). By Lemma 4.19, we can lift the factorization systems, so we are done. □

Proof. By pushing out the initial maps to the two factors \(A\), we see that each of the inclusions into \(A\cup A\) is a cofibration, so the components of \(i\) are as well. Moreover, since the map \(\Delta ^1A \to A\) is a weak equivalence, by the \(2\) out of \(3\) property, so are the components of the inclusion.

Reflexivity of left homotopy comes from the canonical map \(\Delta ^1A \to A\). Symmetry comes from swapping the factors. Transitivity follows by observing that we can pushout two left homotopies and two cylinders along a common map to get another homotopy and cylinder. We need \(A\) to be cofibrant so that the pushout is also a cylinder. □

Proof. It will suffice to show by duality that if we have a right homotopy, and a cylinder object \(\Delta ^1A\) for \(A\), then \(f,g\) are left homotopic with respect to \(\Delta ^1A\). Let \(p_0,p_1\) be the components of the projection \(B\Delta ^1 \to B\times B\), let \(\sigma : B\to B\Delta ^1\) be the equivalence coming from the path structure, and let \(h\) be the right homotopy. Then the map \(p_1\gamma \) in the diagram below is a left homotopy.

□

Proof. Since homotopy equivalences are closed under composition and we can factor a weak equivalence into a trivial fibration and cofibration, it suffices by duality to show it when \(f\) is a trivial cofibration. If \(\cdot \) denotes the terminal object, let \(g\) denote a lift of the diagram

By construction it is a left inverse so it suffices to show \(f \circ g\) is homotopic to the identity. This can be achieved by finding a lift in the diagram below.

□

Proof. It is ok to just think about the full essentially surjective subcategory of the homotopy category consisting of cofibrant fibrant objects. Now suppose we have a functor \(F:C \to D\) inverting weak equivalences. Since the two inclusions into a cylinder are equal in the homotopy \(1\)-category, \(F\) factors through the category obtained by identifying homotopy classes of maps. By Lemma 4.24, weak equivalences are already isomorphism in this category, so the factorization is unique.

Now suppose that a map \(f:X \to Y\) is an isomorphism in the homotopy category. After passing to cofibrant fibrant replacements using the \(2\) out of \(3\) property, we can assume that our objects are cofibrant fibrant, so that \(f\) has a homotopy inverse \(g\). Moreover, by factoring \(f\) into a trivial cofibration and fibration and using Lemma 4.24, we can assume \(f\) is a fibration. The strategy is to show that \(f\) has the right lifting property with respect to cofibrations. Let \(\Delta ^1X,\tilde {Y}\) be path objects for \(X,Y\), \(k\) a right homotopy from \(gf\) to the identity, and form the diagram below given a map from a cofibration \(i\) to \(f\) given by maps \(\alpha ,\beta \):

If we could fill in the map \(?\) in a way that \((p_0,?)\) is a fibration and the diagram commuted, it would have to be a trivial fibration by the \(2\) out of \(3\) property for the composition with \(X\times _{\pi _2} Y \to X\), so we could make the indicated lift and be done.

To make \(?\) we will choose a particularly nice right homotopy from \(gf\) to the identity by slightly modifying the dual construction of Proposition 4.23.

Let \(h:\Delta ^1X \to X\) be a left homotopy from \(gf\) to the identity, and \(s\) the map \(\Delta ^1X \to X\). Choose a path object \(Y\Delta ^1\) for \(Y\). We can make a path object \(X\Delta ^1\) for \(X\) with a compatible fibration \(\hat {f}=?\) to \(Y\Delta ^1\) by factoring the map \((\Delta ,\sigma _Y f): X \to (X\times X)\times _{Y\times Y} Y\Delta ^1\) as a trivial cofibration and fibration. Then create a lift \(Q\) in the diagram below.

\(k=i_1Q\) is a right homotopy from \(gf\) to the identity with the extra property that \(\hat {f}k = \sigma _Yf\). This shows that the strategy works. □

Proof. Let \(X\) be cofibrant, and \(Y\) be fibrant. We can choose a cofibrant replacement \(Y'\) with a trivial fibration to \(Y\). The induced map on homotopy classes of maps from \(X\) is surjective since \(X\) is cofibrant. By looking at the map \(X \cup X \to \Delta ^1X\) in \(C_{X\cup X/}\), it is injective on homotopy classes. Making the dual argument we can reduce to the case when \(X,Y\) are cofibrant and fibrant, where we can use Theorem 4.25. □

Proof. Choose \(Y\Delta ^1\) with structure map \(\sigma \), and a correspondence \(\gamma \) between \(h\) and a right homotopy \(k\) on \(Y\Delta ^1\). \(\gamma \) glues on \(\overline {\Delta ^1}X\cup _{X\cup X}\overline {\Delta ^1}X\) to give a map \(\overline {\gamma }\), so we can lift the diagram

and the map \(p_1K=H\) works. □

Proof. \(f\) admits a section \(g\) by lifting the identity along \(f\), and since this is an inverse in the homotopy category, there is a left homotopy \(h: \Delta ^1X \to X\) from the identity to \(gf\). Let \(h^{-1}\) denote tha homotopy in the other direction obtained from swapping factors. \(gfh^{-1}\) is a homotopy from \(gf\) to itself, and let \(gfh^{-1}\circ h:\overline {\Delta ^1}X \to X\) be the composite of the homotopies \(gfh^{-1}\) and \(h\).

By Lemma 4.28 \(f(gfh^{-1}\circ h) = fh^{-1}\circ fh\) is homotopic to the identity via some homotopy \(H:X' \to Y\), where \(i_L,i_R\) are the inclusions \(\overline {\Delta ^1}X \to X'\), we can create a lift \(K\) as below so that \(Ki_R\) will be the desired fibrewise homotopy.

□

Proof. By Proposition 4.15, it suffices to show if \(f \in D\) has the left lifting property with respect to \(g\) such that \(Uf\) is a fibration, then \(Uf\) is a weak equivalence. Now produce a lift \(u\) of the diagram

Then construct a lift in the diagram

, so that in the diagram

The upper triangle commutes and the lower triangle does up to right homotopy. Applying \(U\) to this square, which preserves right homotopy, we see that \(UA \to UB\) is an isomorphism in the homotopy \(1\)-category of \(C\). □

Proof. Choose a homotopy \(A\cup A \to \Delta ^1A \to C\) between \(g'f\) and \(i\) where \(g'\) is any map whose homotopy class is \(\bar {g}\). We can choose a compatible cylinder \(\Delta ^1A\coprod _{A\cup A} (B\cup B) \to \Delta ^1B \to B\) by factoring the natural map to \(B\) as a cofibration and a trivial fibration.

The homotopy naturally extends to \(\Delta ^1A \coprod _{A} B \to C\), and \(\Delta ^1A\coprod _AB \to \Delta ^1B\) is a trivial cofibration, so since \(X\) is fibrant, it extends to \(\Delta ^1B\), giving a homotopy between \(g'\) and a \(g\) solving the problem. □

Proof. We will show that \(j'\) is an equivalence in the homotopy category. Let \(X\) be an object, and consider the map \(\Hom _{hC}(B',X) \to \Hom _{hC}(B,X)\). To see it is surjective, observe that given a map \(f:B \to X\), we can homotope its restriction to \(A\) to something restricted from \(A'\). By Proposition 4.32, the original map can be homotoped to exhibit \(f\) as in the image.

To see injectivity, if \(s, s'\) are two identified maps, then there is a commutative square

By the surjectivity statement in the category \(C_{/X\times X}\), there is a lift in this square, giving injectivity.

□

Proof. This follows immediately from Lemma 4.33. □

Proof. Suppose that there are two cofibrant pushout diagrams with a weak equivalence to a homotopy pushout square. We can find an equivalent cofibrant pushout diagram mapping to each of them in a compatible way as follows: Take something mapping compatibly into both of the top left objects of the pushout squares, replace it with an equivalence from a cofibrant object \(X\). The same method in \(C_{/X}\) yields an entire cofibrant pushout square mapping to each.

Thus it suffices to show that for any two cofibrant pushout squares, the pushout is an equivalence. Using Lemma 4.33, we can reduce to the case that we are only changing the object in the upper left corner: if we change only one of the other objects by an equivalence, that Lemma along with gluing for pushout squares implies the result.

Furthermore, we can factor the induced map on the upper left vertex into a trivial cofibration and trivial fibration. In the case of a trivial fibration, there is a section since the objects are cofibrant, so picking a section, we there is another pushout square between the diagonals of the pushout squares, and we are done using Lemma 4.33 again. Thus we can assume that the map in the upper left hand corner is a trivial cofibration, in which case it has a retraction. This induces a retraction on the pushout, showing that the pullback map is surjective on homotopy classes of maps to any space. By again mapping to \(X\Delta ^1\) relative to \(X\times X\), we see it is injective as well on homotopy classes, so the map on pushouts is an equivalence. □

Proof. Take a cofibrant replacement of the pushout square, and observe that it induces an equivalence on the pushout by either using Lemma 4.33 under the first hypothesis, or the fact that \(C\) is left proper. □

Proof. WLOG, we can assume that the diagram is a pullback square of fibrant objects and the maps \(E'\to B\), \(E\to B\) are fibrations. Then by Proposition 3.4, there are long exact sequences associated to both fibrations, and the fibres can be identified. We can view the map between the long exact sequences as a double complex with two nontrivial rows. Since the rows are exact, the total complex is exact. But the total complex is the Mayer-Vietoris long exact sequence with an extra copy of \(\pi _n\) of the fibres that can be removed since the fibres are just identified. □

Proof. Say an ordinal is even if it is a limit ordinal plus an even number, and odd otherwise. Let \((\kappa )\) be the poset of ordinals of size \(<\kappa \) and let \((\kappa )_0\) and \((\kappa )_1\) denote the even and odd ones respectively.

Suppose we can construct a commutative diagram

Then since \(\kappa _0\) and \(\kappa _1\) are cofinal in \((\kappa )\), their colimits in \(D\) are isomorphic, but this agrees with the image of their colimits in \(C,C'\).

We can construct such a diagram inductively. On limit ordinals, we can extend by taking the colimit in either \(C\) or \(C'\). On sucessor ordinals, we can extend because we assumed that \(p,p'\) are \(\kappa \)-cofinal. □

Proof. Let \(P_A,P_B\) be tha poset of \(\tau \)-small \(\kappa \)-filtered subsets of \(A\) (resp \(B\)) which contain \(A_0\) (or \(B_0\)).Let \(X\) be the common colimits and consider the maps \(P_A,P_B \to D_{/X}\) given by taking colimits. Then applying Lemma 4.42 to this situation, we are done. □

Proof. Choose \(\tau \) large enough so that \(C\) is \(\tau \)-presentable, let \(C^{\tau }\) be the \(\tau \)-compact objects. We’ll see that \(C^{\tau }\cap C_0\) generates \(C_0\) under \(\kappa \)-filtered colimits. Given an object in \(C_0\), present it as a colimit of a filtered diagram \(A\). Let \(S\) be the collection of \(\tau \)-small \(\kappa \)-filtered subdiagrams of \(A\) such that the colimit is in \(C_0\). We can then take the colimit over \(S\) of the colimits of subdiagrams of \(A\) to obtain the same object.

Conversely, suppose that \(C\) is generated under \(\kappa \)-filtered colimits by a small subcategory \(D \subset C_0\). Choose \(\tau \gg \kappa , \tau > \kappa \) such that every object of \(D\) is \(\tau \)-compact. To see that the condition is satisfied, suppose that we have a diagram \(X_{\alpha }, \alpha \in A\), with \(X_A \in C_0\) and \(A\) \(\tau \)-filtered. \(X_A\) is also the colimit over \(Y_{\beta }, \beta \in B\) with \(B\) a \(\tau \)-small \(\kappa \)-filtered diagram in \(D\). By replacing \(B\) with the family of \(\tau \)-small \(\kappa \)-filtered subdiagrams of \(B\), and using the family \(Y_{B'}\) for all such subsets, we can assume that \(B\) is \(\tau \)-filtered. Now we can apply Lemma 4.43. □

Proof. The subcategory of \(\Set ^{[1]}\) consisting of surjections is accessible (everything is a \(\kappa \)-filtered colimit of finite surjections). For any morphism \(i:a\to b\), the functor \(A^{[1]} \to \Set ^{[1]}\) sending \(f:x \to y\) to \(\Hom (b,x) \mapsto \Hom (b,y)\times _{\Hom (a,y)}\Hom (a,x)\) preserves filtered colimits if \(A,B\) are \(\kappa \)-compact.

\(F\) and \(W\cap F\) are obtained as the preimage of surjections on products of such maps (since they are defined via lifting properties with respect to a set of maps). Thus by Corollary 4.45 they are accessible.

By Proposition 4.8, there is a functorial factorization into a trivial cofibration and fibration. \(W\) is exactly the preimage of those that factor as a trivial cofibration and trivial fibration, so is accessible too. □

Proof. Choose \(\kappa \) so that \(W\) is \(\kappa \)-accessible, choose \(\tau \gg \kappa , \tau > \kappa \) such that Lemma 4.44 holds, \(C\) is \(\tau \)-accessible, and everything in \(C_0\) is \(\tau \)-compact. By Lemma 2.3, we can replace \(C_0\) with \(C \cap A^{\tau }\).

Let \(S = C_0 \cap W\). We will show that \(S\) generates \(C \cap W\) as a weakly saturated class of morphisms. Let \(\overline {S}\) be the weak saturation of \(S\). Suppose that \(f: X \to Y\) is a morphism in \(C \cap W\). By Corollary 2.77, there is a \(\tau \)-good \(C_0\)-tree \(\{Y_{\alpha }\}_{\alpha \in A}\) with root \(X\) such that \(Y\) is isomorphic to \(Y_A\) as objects in \(C_{X/}\). Say that a subset \(B\) of \(A\) is good if it is downward-closed and the canonical map \(X \to Y_B\) is in \(W\). We would like to prove that \(A\) is good.

The set of good \(A\) is closed under transfinite union.

Suppose that \(B,B' \subset A\) are such that \(B,B', B\cap B'\) are good. Then \(B\cup B'\) is good. To see this, we consider the pushout

The morphisms are all in \(C\), and the upper horizontal map is in \(W\) by \(2\) out of \(3\). Since \(C\cap W\) is closed under pushouts, the lower horizontal map is in \(W\) too, so by \(2\) out of \(3\) again, the map \(X \to Y_{B'} \to Y_{B' \cup B}\) is in \(W\).

Next, observe that if \(B_0\subset A\) is \(\tau \)-small, then there is a \(\tau \)-small \(B\supset B_0\) such that \(Y_{B_0}\) is in \(C_0\), and so that \(B\) is good.

Next we will prove:

• • Let \(A'\) be a good subset of \(A\) and let \(B_0\subset A\) be \(\tau \)-small. Then there exists a \(\tau \)-small subset \(B_0 \subset B\subset A\) with \(B,B\cap A'\) good.

Start by setting \(B'_0 = A \cap B_0\). We will inductively construct increasing sequences of \(\tau \)-small subsets \(B_i,B'_i\) as follows: Choose \(B_{i+1}\) to be a \(\tau \)-small good subset of \(A\) containing \(B_i \cup B'_i\) and \(B'_{i+1}\) to be a \(\tau \)-small good subset of \(A\) containing \(B_{i+1} \cap A'\). Taking the union over \(i\), we will have proven the desired result.

Now there is no obstruction to making a good set larger, so \(A\) is good by Zorn’s Lemma. □

Proof. A presentable category has finite limits and colimits. To see we get a model structure, we need to check that \((C,(C\cap W)^{\oslash })\) and \((C\cap W,C^{\oslash })\) are weak factorization systems, which follows from Proposition 4.8. To see it is combinatorial, we just observe that by Lemma 4.47, \(C\cap W\) is generated by a small set in addition to \(C\). □

Proof. First we will show left properness. First observe that \((2)\) actually holds for \(f\) in a weakly saturated set (this uses that \(W\) is stable under filtered colimits), so it holds for \(C\), the weak saturation of \(C_0\).

To check the conditions of Corollary 4.48, we only need to check that \(C\cap W\) is weakly saturated. It is stable under retracts since \(W\) is stable under filtered colimits. To see that \(C\cap W\) is stable under pushouts, we can factor any pushout as a pushout by a trivial fibration and a cofibration as below.

We then know by left properness that \(f' \in C\cap W\) and hence again by left properness \(h' \in W\). By the two out of \(3\) property \(f'' \in W\). □

Proof. Let \(h_C\) denote the localization functor of \(C\). Since \(F\) is homotopical on \(C_Q\), \(h_DFQ\) does descend to a functor from \(h(C)\). \(h_DFq\) gives the natural transformation we want. Instead of proving the universal property in \(h(D)^C\), it suffices to prove it in the equivalent full subcategory of \(h(D)^C\) consisting of homotopical functors. Now suppose that \(G\in h(D)^C\) is homotopical and we have a natural transfromation \(\eta :G\to h_DF\). \(Gq:GQ\to G\) is a natural transformation that is a natural isomorphism since \(G\) is homotopical. Thus by naturality of \(\eta \), \(\eta \) factors as \(h_DF q \circ \eta Q \circ (Gq)^{-1}\). Uniqueness follows from the fact that \(h_DFq\) is a natural isomorphism since \(q\) is a weak equivalence. □

Proof. Let \(f:X \to Y\) be a weak equivalence of cofibrant objects. Factor \(X \coprod Y \xrightarrow {f,1} Y\) as a cofibration and trivial cofibration, and consider the diagram below.

By \(2\) out of \(3\), \(i,j\) are weak equivalences, so they get sent to weak equivalences. But then since the map \(Z \to Y\) is a retraction of \(i\), it is also sent to a weak equivalence by \(2\) out of \(3\). Then \(f\) is too by \(2\) out of \(3\). □

Proof. The first fact follows from 4.58 and second just requires the additional observation that the inclusion of the full subcategory of cofibrant objects induces an equivalence on the homotopy category. □

Proof. The preserving limits and colimits is satisfied because of the adjunction. The equivalences then follow because of Lemma 2.13. □

Proof. Suppose \(F\dashv G\) is a Quillen equivalence. Then for \(c\) cofibrant and \(d\) fibrant, \(F,G\) agree with the derived functors, so combined with the fact that weak equivalences are the isomorphism in the homotopy category, the condition follows. Conversely if the condition holds, then the unit map \(c \to GFc\) is an isomorphism in \(h(C)\) since \(G\) is given by taking a fibrant replacement and applying \(G\), so by the condition it is equivalent to \(Fc\) being isomorphic or weakly equivalent to the fibrant replacement. □

Proof. These claims follow from the condition on \(\hat {\otimes }\) and Lemma 4.58. □

Proof. This follows from Lemma 2.27. □

Proof. We can replace all the model categories with the subcategory of cofibrant objects. The left Quillen functors are still defined since they take cofibrations to cofibrations and the initial object to the initial object. Now there is no deriving necessary. □

Proof. Produce a diagram

by choosing a fibrant replacement for \(Y\) and factoring \(X \to Y \to RY\). The map \(Y\times _{RY}RX \to RX\) is an equivalence by hypothesis, so \(X \to Y\times _{RY}RX \) is an equivalence by \(2\) out of \(3\). Pulling back along \(Z \to Y\) is a right Quillen functor, so it preserves weak equivalences between fibrant objects. Applying this to \(X \to Y\times _{RY}RX \), we get the desired result. □

Proof.

Suppose that \(C\) is a simplicial model category. Clearly \((1)\) and \((2)\) are satisfied, as well as \(3\), since the map \(\Delta ^n \to \Delta ^0\) admits a section which is a trivial fibration.

Now suppose the conditions are satisfied. Since every object is cofibrant, \(C\) is left proper, and moreover the functor \(\otimes X:C \to C\) is homotopical.

We will first see that the tensoring bifunctor \(C\times \SSet \to C\) preserves weak equivalences in each variable. Fix \(c \in C\) and suppose that \(K \to K'\) is a weak equivalence in \(\SSet \). Factor it as a trivial fibration and trivial cofibration. The trivial fibration has a section, which is a trivial cofibration, (since it is injective) so we can reduce to the case of \(K \to K'\) being a trivial cofibration. The collection of all such \(K \to K'\) such that \(c\otimes K \to c\otimes K'\) is a trivial cofibration is weakly saturated, so we reduce to showing it for the horn inclusions \(\Lambda ^n_i \to \Delta ^n\). The inclusion of a vertex \(\{v\} \to \Lambda ^n_i\) is a pushout of horn inclusions of dimension \(<n\), so by induction and the \(2\) out of \(3\) property it suffices to show it for \(\{v\} \to \Delta ^n\) which was assumed.

Next fix \(K\), and we will see that \(c \to c\otimes K\) preserves weak equivalences. Let \(f:c \to c'\) be a weak equivalence. Since weak equivalences are closed under filtered colimits, it suffices to show this for \(K\) finite. We will then prove it by induction on the subcomplex of \(K\). Choose a subcomplex \(L\) of \(K\) for which the result is satisfied, and a nondegenerate cell \(\sigma \) we would like to adjoin. \(f\otimes \partial \sigma , f\otimes L\) are equivalences by the inductive hypothesis, and \(f\otimes \sigma \) is an equivalence by assumption.

Because our model structure is left proper, using assumption \((2)\) we get that \(D\otimes (-)\) applied to the diagram adjoining \(\sigma \) to \(L\) is a homotopy pushout for any \(D\). Thus since each of the components is a weak equivalence, the pushout is a weak equivalence.

Suppose now that \(i:c \to c'\) and \(j:S \to S'\) are cofibrations in \(C\) and \(\SSet \) respectively. We would like to prove that the induced map \(i \wedge j: (c\otimes S') \cup _{(c\otimes S}(c'\otimes S')\) is an equivalence if either \(i,j\) are equivalences. But this follows from applying \(2\) out of \(3\) and left properness to the diagram

□

Proof. A lift in a diagram of the form

is the same as a lift of a \(0\)-simplex in the map \(\Hom (B,X) \to \Hom (A,X) \times _{\Hom (B,Y)}\Hom (X,Y)\). But this is a trivial fibration if either of the vertical maps is trivial by the fact that the two-variable adjunction with \(\otimes \) is Quillen, so there is a unique lift up homotopy. □

Proof. It is clear that if \(f\) is an equivalence, then the other map is an equivalence. Conversely, if the other map is an equivalence, since \(\ul \Hom (B,C) \to \ul \Hom (A,C)\) is a fibration, it is a trivial fibration, and in particular bijective on \(\pi _0\). Thus \(f\) is an isomorphism in the homotopy category. □

Proof. By Proposition 4.15 a model structure, and we only need to show that it is simplicial.

To do this, we can identify the map

\[U\hom _{D}(L,X) \to U(\hom _{\simp C}(K,X)\times _{\hom _{D(K,Y)}} \hom _{D} (L,Y) )\]

with

\[\hom _{C}(L,UX) \to \hom _{C}(K,UX)\times _{\hom _{C(K,UY)}} \hom _{C} (L,UY)\]

since the adjunction is enriched and \(U\) preserves pullbacks. If \(K \to L\) is a cofibration and \(X \to Y\) is a fibration, one of which is trivial then this map is a fibration, since \(C\) is a simplicial model category. □

Proof. We have a functorial path object given by \(\hom (\Delta ^1,-)\), and \(Q\) is a functorial fibrant replacement, so by Proposition 4.30 we are done. □

Proof. \(\simp {C}\) is tensored and cotensored by Lemma 2.29. The adjunction \(\adjunction {F}{\simp C}{\SSet }{U}\) enriches over \(\SSet \) by Lemma 2.28 since \(F(X \times K) \cong F(X)\otimes K\). \(F\) sends the generators of cofibrations/trivial cofibrations to \(\kappa \)-small objects after possibly enlarging \(\kappa \) since \(\Delta ^{op}\) is a small category, so by Proposition 4.78 we are done. □

Proof. Use \(\Ex \) to produce a natural fibrant replacement...[INCOMPLETE] □