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\begin{document}
\title{Model Categories}
\author{Lennart Meier \& Irakli Patchkoria}
\date{}
\maketitle
Model categories are a way of abstracting homotopy theory. Whenever one has abstractions, one must have examples.
\section{Examples}
\begin{example}[Classical homotopy theory (\cat{Top})]
There are two kinds of equivalences: homotopy equivalences and weak homotopy equivalences (for which definition we require $f_*$ to be an isomorphism for all basepoints in the domain). We then have the famous
\begin{theorem}[Whitehead]: A weak equivalence of cell complexes is actually a homotopy equivalence. \end{theorem}
Here, a cell complex is slightly more general than a CW-complex. A \textit{relative cell complex} is as follows. Suppose $A\subset X$. Then $(X,A)$ is a relative cell complex is there a sequence of spaces $X_n$ such that $X_0=A$ and
\begin{diagram}
\coprod S^{i_n-1} & \rTo & X_{n_1} \\
\dTo & & \dTo \\
\coprod D^{i_n} & \rTo & X_n
\end{diagram}
is a pushout square. (The generalization here is that smaller cells can be glued to higher cells.) Then we require $X=\colim \ X_n$.
An important fact is that every space is weakly equivalent to a cell complex. Moreover, let $(B,A)$ be a relative cell complex and $f:X\ra Y$ be a Serre fibration and weak equivalence. Then in every commutative diagram of the form
\begin{diagram}
A & \rTo & X \\
\dTo & \ruDashto^l & \dTo_f \\
B & \rTo & Y
\end{diagram}
there is a lift $l:B\ra X$ making the two triangles commutative. (Recall that $f:X\ra Y$ is a \textit{Serre fibration} if we always have
\begin{diagram}
D^n & \rTo & X \\
\dTo & \ruDashto & \dTo_f \\
D^n\times I & \rTo & Y.
\end{diagram}
These have long exact sequences in homotopy groups.)
We define the \textit{homotopy category of topological spaces} $Ho(\cat{Top})$ to have $obj(Ho(\cat{Top}))=obj(\cat{Top})$ with morphisms the homotopy classes of maps: $Ho(\cat{Top})(X,Y)=[QX,QY]$, where $QX$ is a cell complex weakly homotopy equivalent to $X$. (This definition does not depend on the choices of $QX$ and $QY$, because it does not change the set of homotopy classes of maps.)
\end{example}
\begin{example}[Chain complexes of $R$-modules bounded below by 0: $Ch_{\geq 0}(R)$]
There are again two notions of equivalence: a chain map $f:X\ra Y$ is a \textit{chain homotopy equivalence} if there exists a chain map $g:Y\ra X$ such that $fg\simeq id$ and $gf\simeq id$. Recall that $f,f':X\ra Y$ are \textit{chain homotopic} (and we write $f\simeq f'$) if there exists a map $H:X_*\ra Y_{*+1}$ such that $H\partial +\partial H = f-f'$.
We can write this differently. Let $X$ and $Y$ be chain complexes. Define $(X\otimes Y)_n = \oplus_{i+j=n} X_i \otimes Y_j$ with $\partial(x\otimes y)=\partial x\otimes y + (-1)^{|x|}x\otimes \partial y$. Then define the interval complex $R \ra R\oplus R$ (in degrees 1 and 0) with differential given by $(1,-1)$. Then a chain map is just given by $\overline{H}:I\oplus X\ra Y$.
The second kind of equivalence is called a \textit{quasi-isomorphism}. $f:X\ra Y$ is a quasi-isomorphism if it induces isomorphisms $f_*:H_nX\ra H_nY$. Obviously every chain homotopy equivalence is a quasi-isomorphism. Moreover, we have the following
\begin{theorem} If $X,Y$ are complexes of projectives, then $f$ is a quasi-isomorphism iff it is a chain homotopy equivalence. \end{theorem}
One can similarly define the homotopy category of chain complexes. This is often called the \textit{derived category}. $\mc{D}^{\geq 0}(R)=Ho(Ch_{\geq 0}(R))$ has objects those of $Ch_{\geq 0}R$, and $D(R)(X,Y)=[QX,QY]$ where $QX$ is a replacement of $X$ by a module of projectives.
\end{example}
\section{Model Categories}
We now abstract the common features of these two examples to the notion of a model category. Let $\mc{C}$ be a category admitting all small limits and colimits. Then a \textit{model structure} on $\mc{C}$ consists of three classes of morphisms: weak equivalences (in our examples, weak homotopy equivalences / quasi-isomorphisms), cofibrations (relative cell complexes / split inclusions (i.e.\! maps where the cokernel only consists of projectives)), and fibrations (Serre fibrations / degreewise epimorphism above degree 0). These must satisfy some axioms:
\begin{enumerate}
\item \textit{(the 2 out of 3 property)} If $f:X\ra Y$ and $g:Y\ra Z$ are morphisms then if two of $f$, $g$, and $gf$ are w.e.'s then the third is too.
\item \textit{(retract axiom)} In the diagram of retracts
\begin{diagram}
X & \rTo & Y & \rTo & X \\
\dTo^g & & \dTo^f & & \dTo_g \\
Z & \rTo & W & \rTo & Z,
\end{diagram}
if $f$ is a weak equivalence / cofibration / fibration, then then same holds for $g$.
\item \textit{(lifting axiom)} Consider the diagram
\begin{diagram}
A & \rTo & X \\
\dEmbed^f & & \dOnto_g \\
B & \rTo & Y
\end{diagram}
where $f$ is a cofibration and $g$ is a fibration. Then there must be a lift $l:B\ra X$ making the two triangles commute if either $f$ or $g$ is a weak equivalence.
\item \textit{(factorization axiom)} Let $f:X\ra Y$ be a map. Then there exist a factorization $p\circ i:X\ra Z\ra Y$ such that $i$ is a cofibration and $p$ is a fibration, and we can choose either $i$ or $p$ to be a weak equivalence.
\end{enumerate}
We call a \textit{trivial fibration} a fibration which is also a weak equivalence. Similarly, we call a \textit{trivial cofibration} a cofibration which is also a weak equivalence.
We say that cofibrations have the \textit{left lifting property} w/r/t all trivial fibrations, and dually all fibrations have the \textit{right lifting property} w/r/t all trivial cofibrations.
This corresponds to the lifting theorem we wrote down in the case of topological spaces.
\begin{example} Here are a few examples of model categories:
\begin{itemize}
\item topological spaces with weak homotopy equivalences, retracts of relative cell complexes, and Serre fibrations
\item chain complexes $Ch_{\geq 0}(R)$ with qi's, maps that are injective and have projective cokernel in each degree, maps that are surjective in every degree above 0
\item simplicial sets
\item diagram categories
\item algebras over topological operads
\item $\ldots$
\end{itemize}
\end{example}
\begin{remark} We denote an initial object by $\emptyset$ and a terminal object by $\ast$. Then for any space $X$ we can apply the factorization axiom to the unique maps: $\emptyset\ra X$ gives $\emptyset \cofib QX \triv{\fib} X$. If $\emptyset \ra Z$ is a cofibration, then $Z$ is called \textit{cofibrant}. Similarly, $X\ra \ast$ factorizes to $X \triv\cofib RX \fib X$. Then $QX$ is called the \textit{cofibration replacement} of $X$ and $RX$ is called the \textit{fibrant replacement} of $X$. [Note that in $\cat{Top}$, all objects are fibrant.] \end{remark}
\begin{proposition}
Every map having the LLP with respect to all trivial fibrations is a cofibration. (Dually, every map having the RLP with respect to all trivial cofibrations is a fibration.)
\end{proposition}
\begin{proof}
Let $f:X\ra Y$ be a map with the LLP w/r/t trivial fibrations. Factorize $X \cofib Z \triv\fib Y$. Then we get the diagram
\begin{diagram}
X & \rTo & Z \\
\dTo & \ruDashto_h & \dOnto \\
Y & \rEquals & Y.
\end{diagram}
This gives us the diagram
\begin{diagram}
X & \rEquals & X & \rEquals & X \\
\dTo_f & & \dEmbed_i & & \dTo_f \\
Y & \rTo_h & Z & \rTo_p & Y
\end{diagram}
using the retract axiom. Thus $f$ is a cofibration.
\end{proof}
In a general model category we define a \textit{cylinder object} for $B\in \mc{C}$ to be a factorization of $B\coprod B \ra B$ as $B\coprod B \cofib Z \stackrel{\sim}{\ra} B$. (One should of course think of this as a cylinder.) Note that we get two maps $i_0,i_1:B\ra Z$ by the universal property of the coproduct. Then, we define a \textit{homotopy} between $f,g:B\ra X$ to be a map $H:Z\ra X$ such that $f=Hi_0$ and $g=Hi_1$. A \textit{homotopy equivalence} is defined as usual.
\begin{proposition}[generalization of Whitehead]
Let $f:X\ra Y$ be a map between cofibrant-fibrant objects. Then $f$ is a homotopy equivalence iff it is a weak equivalence.
\end{proposition}
Predefinition: Define the \textit{homotopy category} $Ho(\mc{C})$ by $obj(Ho(\mc{C}))=obj(\mc{C})$ and $Ho(\mc{C})(X,Y)=[RQX,RQY] = \mc{C}(RQX,RQY)/\sim$. (One can check that $RQX$ is still cofibrant.) We must check 3 things to make sure this makes sense. For cofibrant-fibrant objects:
\begin{enumerate}
\item homotopy equivalence is an equivalence relation;
\item $[RQX,RQY]$ is independent of choices;
\item composition is well defined (i.e. if we have homotopies of two composable maps, we get a homotopy of their composition).
\end{enumerate}
\begin{theorem}
Let $f,g:B\ra X$. Then:
\begin{enumerate}
\item If $f\sim g$ and $h:X\ra Y$, then $hf\sim hg$.
\item If $X$ is fibrant, $h:A\ra B$, then $fh \sim gh$.
\item If $B$ is cofibrant, $~$ is an equivalence relation.
\end{enumerate}
\end{theorem}
Assuming the replacements $R$ and $Q$ are actually functors, then the functor $\mc{C}\ra Ho \ \mc{C}$ given by $X\mapsto RQX$ is the ``minimal category'' where all weak equivalences are sent to isomorphisms.
\section{A taste of the proof that $\cat{Top}$ has a model category structure}
Recall from before that we discussed a model structure on $\cat{Top}$. We have that $p:E\ra B$ is a trivial fibration iff it has a RLP with respect to $\{S^{n-1}\ra D^n\}_{n\geq 0}$. The heart of today's lecture is the following proposition.
\begin{proposition}
Any map $f:X\ra Y$ can be factored as $f=pi$, where $i$ is a cofibration and $p$ is a trivial fibration.
\end{proposition}
There is a very general method for proving such statements, called \textit{Quillen's small object argument}.
\begin{proof}
We will construct inductively the following factorizations:
\begin{diagram}
GAAAAAH
\end{diagram}
We begin with $Z_{-1}=X$. Now suppose we have constructed our diagram through $Z_{n-1}$. Consider the set $\mc{D}$ of all possible commutative diagrams of the form
\begin{diagram}
S^{m-1} & \rTo & Z_{n-1} \\
\dTo & & \dTo \\
D^m & \rTo & Y.
\end{diagram}
We then form the pushout
\begin{diagram}
\coprod_{\infty} S^{m_\alpha -1} & \rTo & Z_{n-1} \\
\dTo & & \dTo \\
\coprod_{\infty} D^{m_\alpha} & \rTo & Z_n. \\
\end{diagram}
The universal property of pushouts then gives us a map $Z_n\ra Y$. Moreover, note that the map $\coprod S^{m_\alpha-1}\ra \coprod D^{m_\alpha}$ is a cofibration, so since the pushout of a cofibration is a cofibration then so is the map $Z_{n-1}\ra Z_n$. So we define $Z=\colim_n Z_n$. Then we get $X\ra Z\ra Y$ factorizing $f$. We know that $X\ra Z$ is a cofibration, and so it only remains to show that $Z\ra Y$ is a fibration.
Now, recall that showing that $p:Z\ra Y$ is a trivial fibration is equivalent to showing that it has the RLP with respect to $\{S^{m-1}\ra D^m\}_{m\geq 0}$. So suppose we have a diagram
\begin{diagram}
S^{m-1} & \rTo & Z \\
\dEmbed & & \dTo_p \\
D^m & \rTo & Y.
\end{diagram}
There is the basic point-set topological fact, coming from the fact that $S^{m-1}$ is compact, that any map $S^{m-1}\ra Z$ factors through some $Z_{n-1}$. So this gives us
\begin{diagram}
& & Z_{n-1} \\
& \ruTo & \dTo \\
S^{m-1} & \rTo & Z \\
\dEmbed & & \dTo_p \\
D^m & \rTo & Y.
\end{diagram}
So now we get the commutative diagram
\begin{diagram}
S^{m-1} & \rTo^\alpha & Z_{n-1} \\
\dTo^{d_0} & & \dTo_{pj_{n-1}} \\
D^m & \rTo & Y
\end{diagram}
which is an element of $\mc{D}$. This gives us
\begin{diagram}
S^{m-1} & \rTo & Z_{n-1} \\
\dTo & & \dTo \\
D^m & \rTo & Z_n.
\end{diagram}
Thus there exists a lift $D^m\ra Z_n$, which when composed with $Z_n\ra Z$ gives us our desired lift $D^m\ra Z$.
\end{proof}
\section{Derived functors \& model categories}
Let $R$ be any ring and suppose $M\in \cat{Mod-}R$ and $N\in R\cat{-Mod}$. Then we get the \textit{Tor groups} $\Tor_n^R(M,N)=H_n(M\otimes_R P_*)$, where $P_*\ra N\ra 0$ is a projective resolution of $N$. We will reformulate this definition in the language of model categories.
The functor $M\otimes_R -:R\cat{-Mod} \ra \Z\cat{-Mod}$ induces a functor $M\otimes_R - :Ch_{\geq 0}(R)\ra Ch_{\geq 0}(\Z)$. We also have the homology functors $H_n:Ch_{\geq 0}(\Z) \ra \cat{AbGrps}$, and we define the composite functor $F_n^\otimes = H_n\circ (M\otimes_R -)$. Thus $\Tor_n^R(M,N)=F_n^\otimes (P_*)$.
In the language of model categories, $P_*\ra N$ is a projective resolution iff $P_*\ra N$ is a trivial fibration and $P_*$ is cofibrant. This motivates the following definition.
\begin{definition}
Let $F:\mc{M}\ra \mc{A}$ be a functor off a model category $\mc{M}$ which sends weak equivalences between cofibrant objects to isomorphisms. Then define the \textit{left derived functor} of $F$ to be the functor $LF:Ho(\mc{M})\ra \mc{A}$ which takes $X$ to $F(QX)$ for any cofibrant replacement $QX \triv\cofib X$.
\end{definition}
Our functor $F_n^\otimes:Ch(R)\ra \cat{AbGrps}$ satisfies the hypothesis of the definition, and so admits a left derived functor $LF_n^\otimes:\mc{D}_{\geq 0}(R)\ra \cat{AbGrps}$, and now we have that $LF_n^\otimes (N)=\Tor_n^R(M,N)$. So $\Tor_n^R(M,-)$ is the composition $R\cat{-Mod} \ra Ch_{\geq 0}(R)\ra \mc{D}_{\geq 0}(R) \ra \cat{AbGrps}$. So the derived functor $LF_n^\otimes$ contains much more information that the Tor groups themselves.
We will indicate why this is independent of cofibrant replacement. This follows from the following lemma.
\begin{lemma}[comparison lemma]
Suppose $f:X\ra Y$ is any map and that $QX$ and $QY$ are any cofibrant resolutions. Then there is a unique $\tilde{f}:QX\ra QY$ up to left homotopy making the diagram
\begin{diagram}
QX & \rOnto^{\sim \ P_X} & X \\
\dDashto^{\tilde{f}} & & \dTo_f \\
QY & \rOnto_{\sim \ P_Y} & Y.
\end{diagram}
commute, and the left homotopy class of $\tilde{f}$ depends only on the left homotopy class of $f$.
\end{lemma}
The easiest thing here is to construct $\tilde{f}$, which we do via the diagram
\begin{diagram}
\emptyset & \rTo & QY \\
\dTo & \ruDashto^{\tilde{f}} & \dTo_{\sim \ P_Y} \\
QX & \rTo_{fP_X} & Y.
\end{diagram}
Thus $f\simeq g$ implies that $F(f)=F(g)$.
\begin{definition}
Suppose we have an adjunction $F:\mc{M}\lras \mc{N}:G$. This is called a \textit{Quillen adjunction} if $F$ preserves cofibrations and trivial cofibrations (or equivalently if $G$ preserves fibrations and trivial fibrations). In this case we call $F$ a \textit{left Quillen functor} and $G$ a \textit{right Quillen functor}.
\end{definition}
The equivalence of the definitions follows from the fact that commutative diagrams of the form
\begin{diagram}
FA & \rTo & X \\
\dTo & \ruDashto & \dTo \\
FB & \rTo & Y
\end{diagram}
are in bijective correspondence to commutative diagrams of the form
\begin{diagram}
A & \rTo & GX \\
\dTo & \ruDashto & \dTo \\
B & \rTo GY.
\end{diagram}
\begin{example}
There is a self-adjunction $F:Ch(\K)\lras Ch(\K):G$ given by $F=C \otimes_\K -$ and $G=\Hom_\K(C,-)$ for a fixed complex $C$. If $C$ happens to be a complex of projectives, then this adjunction is a Quillen adjunction. For example, $\Hom_\K(C,-)$ preserves degreewise surjections.
\end{example}
\begin{example}
Given any model category $\mc{M}$, we have the category $\mc{M}^\omega$ of sequences $X_0\ra X_1 \ra \cdots$ in $\mc{M}$, and there is an adjunction $\colim:\mc{M}^\omega\lras \mc{M}:\textup{const}$ (i.e. $\textup{const}(Y)=Y\ra Y\ra \cdots$). This is a Quillen adjunction.
\end{example}
\begin{example}
There is a Quillen adjunction between $|-|:\cat{sSet}\lras \cat{Top}:S(-)$.
\end{example}
We now have the following central theorem.
\begin{theorem}[Quillen's total derived functor theorem]
Suppose that we have a Quillen adjunction $F:\mc{M}\lras\mc{N}:G$. We can define the composite functors
\begin{diagram}
\mc{M} & \rTo^F & \mc{N} & \rTo^{\gamma_\mc{N}} & Ho(\mc{N}) \\
\mc{N} & \rTo^G & \mc{M} & \rTo^{\gamma_{\mc{M}}} & Ho(\mc{M}).
\end{diagram}
These allow us to construct the derived functors $LF:Ho(\mc{M})\ra Ho(\mc{N})$ and $RG:Ho(\mc{N})\ra Ho(\mc{M})$. These functors are also adjoint.
\end{theorem}
\begin{proof}[Proof sketch]
Suppose (without loss of generality up to isomorphism) that $X\in \mc{M}$ and $Y\in\mc{N}$ are both cofibrant-fibrant. Then
\begin{equation*}
\Hom_{Ho(\mc{M})} (X,RG(Y)) \cong \Hom_{Ho(\mc{M})}(X,G(Y)) = [X,G(Y)] \cong [F(X),Y],
\end{equation*}
where the last isomorphism comes from the fact that Quillen functors respect homotopies (since they respect cylinder objects). Thus the above is
\begin{equation*}
= \Hom_{Ho(\mc{N})}(F(X),Y) = \Hom_{Ho(\mc{N})}(LF(X),Y).
\end{equation*}
\end{proof}
Let us return to our examples. From the self-adjunction on $Ch(\K)$ we get derived tensor product and derived hom. But this is uninteresting, because these are already their own derived functors! More interesting is the Quillen adjunction $\colim:\cat{Top}^\omega\lras\cat{Top}:\textup{const}$. We obtain $L(\colim)$, the \textit{homotopy colimit}, which we can compute (up to homotopy). Take
\begin{diagram}
X=X_0 & \rTo^{f_0} & X_1 & \rTo^{f_1} & X_2 & \rTo^{f_2} & \cdots.
\end{diagram}
Then $L\colim \ X\simeq Tel_nX_n$ is the \textit{mapping telescope}, defined by
\begin{equation*}
Tel_nX_n = \coprod_n X_n \times [0,1] / (x,1)~(f_n(x),0) \textup{ for }x\in X_n.
\end{equation*}
This has that $\pi_k(Tel_nX_n)\cong \colim_n \pi_kX_n$, so the telescope really is the right homotopical notion of the colimit.
Lastly, we will define the correct notion of equivalence.
\begin{definition}
Suppose we have a Quillen adjunction $F:\mc{M}\lras\mc{N}:G$. This induces an adjunction $LF:Ho(\mc{M})\lras Ho(\mc{N}):RG$. We say that $(F,G)$ is a \textit{Quillen equivalence} if $(LF,RG)$ is an adjoint equivalence of categories (i.e. the units and counits of the adjunction are isomorphisms).
\end{definition}
This essentially tells us that the homotopy theories of $\mc{M}$ and $\mc{N}$ are somehow ``the same thing''.
The most important example is that $|-|:\cat{sSet}\lras \cat{Top}:S(-)$ is a Quillen equivalence.
\end{document}