@ -103,16 +103,6 @@ Before we state the uniqueness theorem we need some more properties of minimal m
We have $p^\ast[x]=[px]=0$, since $p^\ast$ is injective we have $x = d \overline{x}$ for some $\overline{x}\in X$. Now $p \overline{x}= y' + db$ for some $b \in Y$. Choose $a \in X$ with $p a = b$, then define $x' =\overline{x}- da$. Now check the requirements: $p x' = p \overline{x}- p a = y'$ and $d x' = d \overline{x}- d d a = d \overline{x}= x$.
\end{proof}
In the following we will need to replace a map by a fibration. But the one given abstractly from the model structure will not fit our needs. So we will first consider the following factorization.\todo{This actually is the same as in chapter 4}
Let $A$ be any cochain complex (not an algebra) and define $C(A)^k = C^k \oplus C^{k-1}$. Then $C(A)$ is again a cochain complex when we define the differential to be $\delta(c_k, c_{k-1})=(0, c_k)$. Note that this cochain complex is acyclic, furthermore there is an obvious surjection $C(A)\tot{\rho} A$. Now for a cochain algebra $A$, we can do the same construction (by forgetting the algebra structure) and apply $\Lambda$. This defines a cdga $\Lambda C(A)$ (which is still acyclic).
Now let $f: X \to Y$ be any map, then we can tensor $X$ with $\Lambda C(Y)$ to obtain:
$$ f: X \tot{x \mapsto x \tensor1} X \tensor\Lambda C(Y)\tot{\psi: x \tensor y \mapsto f(x)\cdot\rho(y)} Y. $$
Where the second map is surjective. By the 2-out-of-3 property the second map is a weak equivalence if and only if $f$ is a weak equivalence. The remarkable thing is that the left map (which is a weak equivalence by the Künneth theorem) has a left inverse, given by $\phi: x \tensor y \mapsto x \cdot\counit(y)$, where $\counit$ is the augmentation.
Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjective and weak equivalences.
\Lemma{minimal-model-bijection}{
Let $f: X \we Y$ be a weak equivalence between cdga's and $M$ a minimal algebra. Then $f$ induces an bijection:
$$ f_\ast: [M, X]\tot{\iso}[M, Y]. $$
@ -120,7 +110,7 @@ Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjec
\begin{proof}
If $f$ is surjective this follows from the fact that $M$ is cofibrant and $f$ being a trivial fibration, see \CorollaryRef{cdga_homotopy_properties}.
By the factorization above, we can turn $f$ into two trivial fibrations (going in different directions). This induces
By \RemarkRef{cdga-mc5a-left-inverse}, we can turn $f$ into two trivial fibrations (going in different directions). Hence we are in the above situation and we find bijections
@ -45,18 +45,33 @@ Note that with these classes, every cdga is a fibrant object.
Now we can find a lift starting at $B$, since $f$ is a cofibration. By precomposition we obtain a lift $B' \to X$.
\end{proof}
Next we will prove the factorization property [MC5]. We will do this by Quillen's small object argument. When proved, we get an easy way to prove the missing lifting property of [MC4]. For the Quillen's small object argument we use classes of generating cofibrations.
Next we will prove the factorization property [MC5]. We will prove one part directly and the other by Quillen's small object argument. When proved, we get an easy way to prove the missing lifting property of [MC4]. For the Quillen's small object argument we use a class of generating cofibrations.
\begin{definition}
Define the following objects and sets of maps:
\begin{itemize}
\item$\Lambda S(n)$ is the cdga generated by one element $a$ of degree $n$ such that $da =0$.
\item$\Lambda D(n)$ is the CDGA generated by two element$b$ and $c$ of degree $n$ and $n+1$ respectively, such that $db = c$ (and necessarily $dc =0$).
\item$I =\{ i_n: \k\to\Lambda D(n)\I n \in\N\}$ is the set of units of $\Lambda D(n)$.
\item$\Lambda D(n)$ is the cdga generated by two elements$b$ and $c$ of degree $n$ and $n+1$ respectively, such that $db = c$ (and necessarily $dc =0$).
\item$I =\{ i_n: \k\to\Lambda D(n)\I n \in\N\}$ is the set of units.
\item$J =\{ j_n: \Lambda S(n+1)\to\Lambda D(n)\I n \in\N\}$ is the set of inclusions $j_n$ defined by $j_n(a)= b$.
\end{itemize}
\end{definition}
\Lemma{cdga-mc5a}{
[MC5a] A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is a trivial cofibration and $p$ a fibration.
}
\Proof{
Consider the free cdga $C =\bigtensor_{x \in X} T(\deg{x})$. There is an obvious surjective map $p: C \to X$ which sends a generator correspondig to $x$ to $x$. Now define maps $\phi$ and $\psi$ in
\[ A \tot{\phi} A \tensor C \tot{\psi} X\]
by $\phi(a)= a \tensor1$ and $\psi(a \tensor c)= f(a)\cdot p(c)$. Now $\psi$ is clearly surjective (as $p$ is) and $\phi$ is clearly a weak equivalence (by the Künneth theorem). Furthermore $\phi$ is a cofibration as we can construct lifts using the freeness of $C$.
}
\Remark{cdga-mc5a-left-inverse}{
The map $\phi$ in the above construction has a left inverse $\overline{\phi}$ given by $\overline{\phi}(x \tensor c)= x \cdot\counit(c)$, where $\counit$ is the natural augmentation of a free cdga (i.e. it send $1$ to $1$ and all generators to $0$). Clearly $\overline{\phi}\phi=\id$, and so $\overline{\phi}$ is a fibration as well.
Furthermore, if $f$ is a weak equivalence then by the 2-out-of-3 property both $\phi$ and $\psi$ are weak equivalences. Applying it once more, we find that $\overline{\phi}$ too is a weak equivalence. So for any weak equivalence $f: A \to X$ we find trivial fibrations $\overline{\phi} : A \tensor C \fib A$ and $\psi: A \tensor C \fib X$ compatible with $f$.
}
\begin{lemma}
The maps $i_n$ are trivial cofibrations and the maps $j_n$ are cofibrations.
\end{lemma}
@ -65,63 +80,49 @@ Next we will prove the factorization property [MC5]. We will do this by Quillen'
\end{proof}
\begin{lemma}
The class of (trivial) cofibrations is saturated.
The class of cofibrations is saturated.
\end{lemma}
\begin{proof}
We need to prove that the classes are closed under retracts (this is already done), pushouts and transfinite compositions. For the class of cofibrations, this is easy as they are defined by the LLP and colimits behave nice with respect to such classes.
However the case of trivial cofibrations does not follow immediately, as we still need to prove that quasi isomorphisms behave as such.\todo{THIS IS HARD}
\end{proof}
As a consequence of the above two lemmas, the class generated by $I$ is contained in the class of trivial cofibrations. Similarly the class generated by $J$ is contained in the class of cofibrations. We also have a similar lemma about (trivial) fibrations.
\begin{lemma}
If $p: X \to Y$ has the RLP w.r.t. $I$ then $p$ is a fibration.
\end{lemma}
\begin{proof}
Let $y \in Y^n$ be an element of degree $n$, then we have the following commuting diagram:
\cdiagram{CDGA_Model_I_Fib}
where $g$ sends the generator $b$ to $y$ and $c$ to $dy$. By assumption there exists a lift $h$. Now $h(b)\in X^n$ is a preimage for $y$, proving that $p$ is surjective.
\end{proof}
As a consequence of the above two lemmas, the class generated by $J$ is contained in the class of cofibrations. We can characterize trivial fibrations with $J$.
\begin{lemma}
If $p: X \to Y$ has the RLP w.r.t. $J$ then $p$ is a trivial fibration.
\end{lemma}
\begin{proof}
\todo{bewijzen}
\end{proof}
We can use Quillen's small object argument with these sets. The argument directly proves the following lemma. Together with the above lemmas this translates to the required factorization. \todo{Definieer wat ``small'' betkent en geef een referentie}
Let $y \in Y$ be of degree $n$ and $dy$ its boundary. By assumption we can find a lift in the following diagram:
\[\xymatrix{
\Lambda S(n+1) \arcof[d]^{j_n}\ar[r]^-{a \mapsto 0}& X \ar[d]^f \\
\Lambda D(n) \ar[r]^-{b \mapsto dy}& Y
}\]
The lift $h: D(n)\to X$ defines a preimage $x' = h(b)$ for $dy$. Now we can define a similar square to find a preimage $x$ of $y$ as follows:
\[\xymatrix{
\Lambda S(n) \arcof[d]^{j_{n-1}}\ar[r]^-{a \mapsto x'}& X \ar[d]^f \\
\Lambda D(n-1) \ar[r]^-{b \mapsto y}& Y
}\]
The lift $h : D(n-1)\to X$ defines $x = h(b)$. This proves that $f$ is surjective. Note that $dx = x'$.
\begin{lemma}
A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is in the class generated by $I$ and $p$ has the RLP w.r.t. $I$.
\end{lemma}
\begin{proof}
This follows from Quillen's small object argument.
Now if $[y]\in H(Y)$ is some class, then $dy =0$, and so by the above we find a preimage $x$ of $y$ such that $dx =0$, proving that $H(f)$ is surjective. Now let $[x]\in H(X)$ such that $[f(x)]=0$, then there is an element $\beta$ such that $f(x)= d\beta$, again by the above we can lift $\beta$ to get $x = d\alpha$., hence $H(f)$ is injective. Conclude that $f$ is a trivial fibration.
\end{proof}
\begin{corollary}
[MC5a] A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is a trivial cofibration and $p$ a fibration.
\end{corollary}
We can use Quillen's small object argument with the set $J$. The argument directly proves the following lemma. Together with the above lemmas this translates to the required factorization. \todo{Definieer wat ``small'' betkent en geef een referentie}
The previous factorization can also be described explicitly as seen in \cite{bousfield}. Let $f: A \to X$ be a map, define $E = A \tensor\bigtensor_{x \in X}T(\deg{x})$. Then $f$ factors as:\todo{This is later defined as $A \tensor\Lambda(C(X))$, which is precisely the same}
$$ A \tot{i} E \tot{p} X, $$
where $i$ is the obvious inclusion $i(a)= a \tensor1$ and $p$ maps (products of) generators $a \tensor b_x$ with $b_x \in T(\deg{x})$ to $f(a)\cdot x \in X$.
\begin{lemma}
\Lemma{cdga-mc5b}{
A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is in the class generated by $J$ and $p$ has the RLP w.r.t. $J$.
\end{lemma}
\begin{proof}
Quillen's small object argument.
\end{proof}
}
\Proof{
This follows from Quillen's small object argument.
}
\begin{corollary}
\Corollary{cdga-mc5b}{
[MC5b] A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is a cofibration and $p$ a trivial fibration.
\end{corollary}
}
\begin{lemma}
[MC4]
\end{lemma}
\Lemma{cdga-mc4}{
[MC4] The lifting properties.
}
\Proof{
One part is already established by definition (cofibrations are defined by an LLP). It remains to show that we can lift in the following situation:
Joshua Moerman
9 years ago