@ -54,7 +54,7 @@ We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A
$$ f_n |_{N(A)_n} : N(A)_n \to B_n. $$
$$ f_n |_{N(A)_n} : N(A)_n \to B_n. $$
Because $f_n$ commutes with the face maps we get:
Because $f_n$ commutes with the face maps we get:
$$ d_i(f_n(x))= f_{n-1}(d_i(x))=0, $$
$$ d_i(f_n(x))= f_{n-1}(d_i(x))=0, $$
for $i>0$ and $x \in N(A)_n$. So the restriction also restircts the codomain, i.e. $f_n |_{N(A)_n} : N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A)\to N(B)$.
for $i>0$ and $x \in N(A)_n$. So the restriction also restricts the codomain, i.e. $f_n |_{N(A)_n} : N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A)\to N(B)$.
Using that $s_i x \in N(X)_{n+1}$ means $0= d_{k+1} s_i x$ for any $k > 0$ and by using using the simplicial equations: $d_{i+1} s_i =\id$, we can conclude $x = d_{i+1} s_i x =0$.
Using that $s_i x \in N(X)_{n+1}$ means $0= d_{k+1} s_i x$ for any $k > 0$ and by using using the simplicial equations: $d_{i+1} s_i =\id$, we can conclude $x = d_{i+1} s_i x =0$.
\end{proof}
\end{proof}
The first lemma tells us that every $n$-simplex in $X$ can be decomposed as a sum of something in $N(X)$ and a degenerate $n$-simplex. The latter lemma asures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:
The first lemma tells us that every $n$-simplex in $X$ can be decomposed as a sum of something in $N(X)$ and a degenerate $n$-simplex. The latter lemma assures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:
\begin{corollary}
\begin{corollary}
\label{cor:NandD}
\label{cor:NandD}
@ -160,12 +160,12 @@ Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$,
again we have written $x$ as $x =\sum_\beta\beta^\ast x_\beta$. Clearly $N(f)= g$. \todo{DK: is this clear?}
again we have written $x$ as $x =\sum_\beta\beta^\ast x_\beta$. Clearly $N(f)= g$. \todo{DK: is this clear?}
\end{proof}
\end{proof}
If we reflect a bit on why the previous functor $C$ was not a candidate for an equivalence, we see that $N$ does a better job. We see that $N$ leaves out all degenerate simplices, so it is more carefull than $C$, which included everything. In fact, corollary~\ref{cor:NandD} exactly tells us $C(X)_n = N(X)_n \oplus D(X)_n$.
If we reflect a bit on why the previous functor $C$ was not a candidate for an equivalence, we see that $N$ does a better job. We see that $N$ leaves out all degenerate simplices, so it is more careful than $C$, which included everything. In fact, corollary~\ref{cor:NandD} exactly tells us $C(X)_n = N(X)_n \oplus D(X)_n$.
\subsection{From $\Ch{\Ab}$ to $\sAb$}
\subsection{From $\Ch{\Ab}$ to $\sAb$}
For the other way around we actually get a functor for free, via abstract nonsense. Let $F : \sAb\to A$ be any functor, where $A$ is an abelian category. We are after a functor $G : A \to\sAb$, this means that if we are given $C \in A$, we are looking for a functor $G(C) : \DELTA^{op}\to\Ab$. Fixing $C$ in the second argument of the $\mathbf{Hom}$-functor gives: $\Hom{A}{-}{C} : A^{op}\to\Ab$, because $A$ is an abelian category. We see that the codomain of this functor already looks good, now if we have some functor from $\DELTA^{op}$ to $A^{op}$, we can precompose, to obtain a functor from $\DELTA^{op}$ to $\Ab$.
For the other way around we actually get a functor for free, via abstract nonsense. Let $F : \sAb\to A$ be any functor, where $A$ is an abelian category. We are after a functor $G : A \to\sAb$, this means that if we are given $C \in A$, we are looking for a functor $G(C) : \DELTA^{op}\to\Ab$. Fixing $C$ in the second argument of the $\mathbf{Hom}$-functor gives: $\Hom{A}{-}{C} : A^{op}\to\Ab$, because $A$ is an abelian category. We see that the codomain of this functor already looks good, now if we have some functor from $\DELTA^{op}$ to $A^{op}$, we can precompose, to obtain a functor from $\DELTA^{op}$ to $\Ab$.
Now recall that we have a family of protoype simplicial sets $\Delta[n]$, which are given by the functor $\Delta : \DELTA\to\sSet$. We can apply the free abelian group pointwise, which gives a functor $\Z^{\ast} : \sSet\to\sAb$. And finally we have our functor $F : \sAb\to A$. Composing these gives:
Now recall that we have a family of simplicial sets $\Delta[n]$, which are given by the functor $\Delta : \DELTA\to\sSet$. We can apply the free abelian group pointwise, which gives a functor $\Z^{\ast} : \sSet\to\sAb$. And finally we have our functor $F : \sAb\to A$. Composing these gives:
$$ F \Z^{\ast}\Delta : \DELTA\to A. $$
$$ F \Z^{\ast}\Delta : \DELTA\to A. $$
We can formally regard this functor as a functor from $\DELTA^{op}$ to $A^{op}$. Now combining this with the $\mathbf{Hom}$-functor gives:
We can formally regard this functor as a functor from $\DELTA^{op}$ to $A^{op}$. Now combining this with the $\mathbf{Hom}$-functor gives:
We've already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets.
We've already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets.
When dealing with homotopy in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for homotopy in simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned about simplicial abelien groups, where there is an obvious choice for the base-point, namely $0$.
When dealing with homotopy in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for homotopy in simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned about simplicial abelian groups, where there is an obvious choice for the base-point, namely $0$.
\subsection{Homotopy groups}
\subsection{Homotopy groups}
\begin{definition}
\begin{definition}
@ -18,9 +18,9 @@ When dealing with homotopy in a topological space $X$ we always need a base-poin
We will call $y$ the \emph{homotopy} and notate $y: x \sim x'$.
We will call $y$ the \emph{homotopy} and notate $y: x \sim x'$.
\end{definition}
\end{definition}
Of course we would like $\sim$ to be an equivalence relation, however this is not true for all simplicial sets. For example there is in general no reason for symmetry, existence of a $1$-simplex $y$ from $x$ to $x'$ does not give us a $1$-simplex $y'$ from $x'$ to $x$. One can give an precise condition on when it is a equivalence relation, the so called \emph{Kan-condition}. In our case of simplicial abelien groups, however, we can prove directly that $\sim$ is an equivalence relation.
Of course we would like $\sim$ to be an equivalence relation, however this is not true for all simplicial sets. For example there is in general no reason for symmetry, existence of a $1$-simplex $y$ from $x$ to $x'$ does not give us a $1$-simplex $y'$ from $x'$ to $x$. One can give an precise condition on when it is a equivalence relation, the so called \emph{Kan-condition}. In our case of simplicial abelian groups, however, we can prove directly that $\sim$ is an equivalence relation.
In figure~\ref{fig:simplicial_htp} it is shown why the definition of homotopy makes sense for $n=1$. Two homotopic $1$-simplices from $Z_n(X)$ are depicted in two ways. The first way only shows the structure we have, indicating what the boundaries are (as described by the face maps). In the second figure we collapsed all occurences of $0$ into a single point. This way of drawing a homotopy should remind the reader of homotopy (between paths) in a topological space.
In figure~\ref{fig:simplicial_htp} it is shown why the definition of homotopy makes sense for $n=1$. Two homotopic $1$-simplices from $Z_n(X)$ are depicted in two ways. The first way only shows the structure we have, indicating what the boundaries are (as described by the face maps). In the second figure we collapsed all occurrences of $0$ into a single point. This way of drawing a homotopy should remind the reader of homotopy (between paths) in a topological space.
Joshua Moerman
12 years ago