@ -11,7 +11,7 @@ We assume the reader is familiar with category theory, basics from algebraic top
\item$\k$ will denote an arbitrary commutative ring (or field, if indicated at the start of a section). Modules, tensor products, \dots are understood as $\k$-modules, tensor products over $\k$, \dots. If ambiguity can occur notation will be explicit.
\item$\cat{C}$ will denote an arbitrary category.
\item$\cat{0}$ (resp. $\cat{1}$) will denote the initial (resp. final) objects in a category $\cat{C}$.
\item$\Hom_\cat{C}(A, B)$ will denote the set of maps from $A$ to $B$ in the category $\cat{C}$. The subscript $\cat{C}$ is occasionally left out if the category is clear from the context.
\item$\Hom_{\cat{C}}(A, B)$ will denote the set of maps from $A$ to $B$ in the category $\cat{C}$. The subscript $\cat{C}$ is occasionally left out if the category is clear from the context.
@ -9,8 +9,12 @@ Given a category $\cat{C}$ and a functor $F: \DELTA \to \cat{C}$, then define th
A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimits. Applying $F$ on these diagrams, make it clear that $F_!$ is functorial. Secondly we see readily that $F^\ast$ is functorial. By using the definition of colimit and the Yoneda lemma (Y) we can prove that $F_!$ is left adjoint to $F^\ast$:
@ -28,8 +32,11 @@ In our case where $F = \Apl$ and $\cat{C} = \CDGA_\k$ we get:
In our case we take the opposite category, so the definition of $A$ is in terms of a limit instead of colimit. This allows us to give a nicer description:
where the addition, multiplication and differential are defined pointwise. Conclude that we have the following contravariant functors (which form an adjoint pair):
Note that $\oint_n(v): \Delta[n]\to\k$ is just a map, we can extend this linearly to chains on $\Delta[n]$ to obtain $\oint_n(v): \Z\Delta[n]\to\k$, in other words $\oint_n(v)\in C_n$. By linearity of $\int_n$ and $x^\ast$, we have a linear map $\oint_n: \Apl_n \to C_n$.
Next we will show that $\oint=\{\oint_n\}_n$ is a simplicial map and that each $\oint_n$ is a chain map, in other words $\oint : \Apl\to C_n$ is a simplicial chain map (of complexes). Let $\sigma: \Delta[n]\to\Delta[k]$, and $\sigma^\ast: \Apl_k \to\Apl_n$ its induced map. We need to prove $\oint_n \circ\sigma^\ast=\sigma^\ast\circ\oint_k$. We show this as follows:
For it to be a chain map, we need to prove $d \circ\oint_n =\oint_n \circ d$. This is very similar to \emph{Stokes' theorem}. \todo{proof this}
We now proved that $\oint$ is indeed a simplicial chain map. Note that $\oint_n$ need not to preserve multiplication, so it fails to be a map of cochain algebras. However $\oint(1)=1$ and so the induced map on homology sends the class of $1$ in $H(\Apl_n)=\k\dot[1]$ to the class of $1$ in $H(C_n)=\k\dot[1]$. We have proven the following lemma.
We now proved that $\oint$ is indeed a simplicial chain map. Note that $\oint_n$ need not to preserve multiplication, so it fails to be a map of cochain algebras. However $\oint(1)=1$ and so the induced map on homology sends the class of $1$ in $H(\Apl_n)=\k\cdot[1]$ to the class of $1$ in $H(C_n)=\k\cdot[1]$. We have proven the following lemma.
\Lemma{apl-c-quasi-iso}{
The map $\oint_n: \Apl_n \to C_n$ is a quasi isomorphism for all $n$.
Joshua Moerman
10 years ago