@ -18,6 +18,58 @@ In the next section homotopy groups of augmented cdga's will be defined. In orde
Define the \Def{augmentation ideal} of $A$ as $\overline{A}=\ker\counit$. Define the \Def{cochain complex of indecomposables} of $A$ as $QA =\overline{A}/\overline{A}\cdot\overline{A}$.
Define the \Def{augmentation ideal} of $A$ as $\overline{A}=\ker\counit$. Define the \Def{cochain complex of indecomposables} of $A$ as $QA =\overline{A}/\overline{A}\cdot\overline{A}$.
}
}
The first observation one should make is that $Q$ is a functor from algebras to modules (or differential algebras to differential modules) which is particularly nice for free algebras, as we have that $Q \Lambda V = V$ for any (differential) module $V$.
The first observation one should make is that $Q$ is a functor from algebras to modules (or differential algebras to differential modules) which is particularly nice for free (differential) algebras, as we have that $Q \Lambda V = V$ for any (differential) module $V$.
\todo{tensor}
The second observation is that $Q$ is nicely behaved on tensor products and cokernels.
\Lemma{Q-preserves-copord}{
Let $A$ and $B$ be two augmented cdga's, then there is a natural isomorphism
\[ Q(A \tensor B)\iso Q(A)\oplus Q(B). \]
}
\Proof{
First note that the augmentation ideal is expressed as
$\overline{A \tensor B}=\overline{A}\tensor B \>+\> A \tensor\overline{B}$
Let $f : A \to B$ be a map of augmented cdga's, then there is a natural isomorphism
\[ Q(\coker(f))\iso\coker(Qf). \]
}
\Proof{
First note that the cokernel of $f$ in the category of augmented cdga's is $\coker(f)= B / f(\overline{A})$ and that its augmentation ideal is $\overline{B}/ f(\overline{A})$. Just as above we make a simple calculation, where $p: \overline{B}\to Q(B)$ is the projection map:
Combining the two lemmas above, we see that $Q$ (as functor from augmented cdga's to cochain complexes) preserves pushouts.
}
Furthermore we have the following lemma which is of homotopical interest.
\Lemma{Q-preserves-cofibs}{
If $f: A \to B$ is a cofibration of augmented cdga's, then $Qf$ is injective in positive degrees.
}
\Proof{
First we define an augmented cdga $U(n)$ for each positive $n$ as $U(n)= D(n)\oplus\k$ with trivial multiplication and where the term $\k$ is used for the unit and augmentation. Notice that the map $U(n)\to\k$ is a trivial fibration. By the lifting property we see that the induced map
is surjective for each positive $n$. Note that maps from $X$ to $U(n)$ will send products to zero and that it is fixed on the augmentation. So there is a natural isomorphism $\Hom_\AugCDGA(X, U(n))\iso\Hom_\k(Q(X)^n, \k)$. Hence
@ -6,18 +6,18 @@ As the eventual goal is to compare the homotopy theory of spaces with the homoto
$$\pi^i(A)= H^i(QA). $$
$$\pi^i(A)= H^i(QA). $$
}
}
This construction is functorial and, as the following lemma shows, homotopy invariant.
This construction is functorial (since both $Q$ and $H$ are) and, as the following lemma shows, homotopy invariant.
\Lemma{cdga-homotopic-maps-equal-pin}{
\Lemma{cdga-homotopic-maps-equal-pin}{
Let $f: A \toB$ be a map of augmented cdga's. Then there is an functorial induced map on the homotopy groups. Moreover if $g: A \to B$ is homotopic to$f$, then the induced maps are equal:
Let $f: A \toX$ and $g: A \to X$ be a maps of augmented cdga's. If$f$ and $g$ are homotopic, then the induced maps are equal:
$$ f_\ast= g_\ast : \pi_\ast(A)\to\pi_\ast(B). $$
$$ f_\ast= g_\ast : \pi_\ast(A)\to\pi_\ast(X). $$
}
}
\Proof{
\Proof{
Let $\phi: A \to B$ be a map of algebras. Then clearly we get an induced map $\overline{A}\to\overline{B}$ as $\phi$ preserves the augmentation. By composition we get a map $\phi': \overline{A}\to Q(B)$ for which we have $\phi'(xy)=\phi'(x)\phi'(y)=0$. So it induces a map $Q(\phi): Q(A)\to Q(B)$. By functoriality of taking homology we get $f_\ast : \pi^n(A)\to\pi^n(B)$.\todo{functoriality is redundant with previous section}
Let $h: A \to\Lambda(t, dt)\tensor X$ be a homotopy. We will, just as in \LemmaRef{cdga-homotopy-homology}, prove that the maps $HQ(d_0)$ and $HQ(d_1)$ are equal, then it follows that $HQ(f)= HQ(d_1 h)= HQ(d_0 h)= HQ(g)$.
Now if $f$ and $g$ are homotopic, then there is a homotopy $h: A \to\Lambda(t, dt)\tensorB$. By the Künneth theorem we have:
Using \LemmaRef{Q-preserves-copord} we can identify the induced maps $Q(d_i) : Q(\Lambda(t, dt)\tensorX)\to Q(X)$ with maps
This means that $f_\ast={d_1}_\asth_\ast={d_0}_\ast h_\ast= g_\ast$. \todo{detail}
Now $Q(\Lambda(t, dt))= D(0)$ and hence it is acyclic, so when passing to homology, this term vanishes. In other words both maps ${d_i}_\ast: H(D(0))\oplus H(Q(A))\to H(Q(A))$ are the identity maps on $H(Q(A))$.
}
}
Consider the augmented cdga $V(n)= S(n)\oplus\k$, with trivial multiplication and where the term $\k$ is used for the unit and augmentation. This augmented cdga can be thought of as a specific model of the sphere. In particular the homotopy groups can be expressed as follows.
Consider the augmented cdga $V(n)= S(n)\oplus\k$, with trivial multiplication and where the term $\k$ is used for the unit and augmentation. This augmented cdga can be thought of as a specific model of the sphere. In particular the homotopy groups can be expressed as follows.
@ -35,4 +35,27 @@ Consider the augmented cdga $V(n) = S(n) \oplus \k$, with trivial multiplication
From now on the dual of a vector space will be denoted as $V^\ast=\Hom_\k(V, \k)$. So the above lemma states that there is a bijection $[A, V(n)]\iso\pi^n(A)^\ast$.
From now on the dual of a vector space will be denoted as $V^\ast=\Hom_\k(V, \k)$. So the above lemma states that there is a bijection $[A, V(n)]\iso\pi^n(A)^\ast$.
\todo{long exact sequence}
In topology we know that a fibration induces a long exact sequence of homotopy groups. In the case of cdga's a similar long exact sequence for a cofibration will exist.
\Lemma{long-exact-cdga-homotopy}{
Given a pushout square of augmented cdga's
\[\xymatrix{
A \ar[d]^-f \arcof[r]^-g \xypo& C \ar[d]^-i \\
B \ar[r]^-j & P
}\]
where $g$ is a cofibration. There is a natural long exact sequence
First note that $j$ is also a cofibration. By \LemmaRef{Q-preserves-cofibs} the maps $Qg$ and $Qj$ are injective in positive degrees. By applying $Q$ we get two exact sequence (in positive degrees) as shown in the following diagram. By the fact that $Q$ preserves pushouts (\LemmaRef{Q-preserves-pushouts}) the cokernels coincide.
Although the abstract theory of model categories gives us tools to construct a homotopy relation (\DefinitionRef{homotopy}), it is useful to have a concrete notion of homotopic maps.
Although the abstract theory of model categories gives us tools to construct a homotopy relation (\DefinitionRef{homotopy}), it is useful to have a concrete notion of homotopic maps.
Consider the free cdga on one generator $\Lambda(t, dt)$\todo{same as $\Lambda D(0)$}, where $\deg{t}=0$, this can be thought of as the (dual) unit interval with endpoints $1$ and $t$. We define two \emph{endpoint maps} as follows:
Consider the free cdga on one generator $\Lambda(t, dt)$, where $\deg{t}=0$, this can be thought of as the (dual) unit interval with endpoints $1$ and $t$. Notice that this cdga is isomorphic to $\Lambda(D(0))$ as defined in the previous section. We define two \emph{endpoint maps} as follows:
$$ d_0, d_1 : \Lambda(t, dt)\to\k$$
$$ d_0, d_1 : \Lambda(t, dt)\to\k$$
$$ d_0(t)=1, \qquad d_1(t)=0, $$
$$ d_0(t)=1, \qquad d_1(t)=0, $$
this extends linearly and multiplicatively. Note that it follows that we have $d_0(1-t)=0$ and $d_1(1-t)=1$. These two functions extend to tensor products as $d_0, d_1: \Lambda(t, dt)\tensor X \to\k\tensor X \tot{\iso} X$.
this extends linearly and multiplicatively. Note that it follows that we have $d_0(1-t)=0$ and $d_1(1-t)=1$. These two functions extend to tensor products as $d_0, d_1: \Lambda(t, dt)\tensor X \to\k\tensor X \tot{\iso} X$.
@ -46,7 +46,7 @@ The results from model categories immediately imply the following results. \todo
\end{itemize}
\end{itemize}
}
}
\Lemma{cdga_homotopy_homology}{
\Lemma{cdga-homotopy-homology}{
Let $f, g: A \to X$ be two homotopic maps, then $H(f)= H(g): HA \to HX$.
Let $f, g: A \to X$ be two homotopic maps, then $H(f)= H(g): HA \to HX$.