Bachelor thesis about the Dold-Kan correspondence
https://github.com/Jaxan/Dold-Kan
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85 lines
6.8 KiB
85 lines
6.8 KiB
\section{Homotopy}
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\label{sec:Homotopy}
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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.
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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 $0$.
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\subsection{Homotopy groups}
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\begin{definition}
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Given a simplicial set $X$ with base-point $\ast$, we define $Z_n(X)$ to be the set of $n$-simplices with the base-point as boundary, i.e.:
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$$ Z_n(X) = \{ x \in X_n | d_i(x) = \ast \text{ for all } i \leq n \}. $$
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For two $n$-simplices $x, x' \in Z_n(X)$, we define $x \sim x'$ if there exists $y \in X_{n+1}$ such that:
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\begin{align}
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d_0(y) &= x \\
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d_1(y) &= x' \\
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d_i(y) &= \ast \text{ for all } i > 1.
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\end{align}
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We will call $y$ the \emph{homotopy} and notate $y: x \sim x'$.
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\end{definition}
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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 Kan-condition. In our case of abelien groups, however, we can prove this directly.
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\todo{Htp: Discuss/picturize Kan-condition?}
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\begin{lemma}
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The relation $\sim$ as defined above is an equivalence relation on $Z_n(X)$. Furthermore it is compatible with addition.
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\end{lemma}
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\begin{proof}
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\emph{Reflexivity}. Let $x \in Z_n(X)$, define $y = s_0 x$. By considering the simplicial identities $d_0 s_0 = \id$ and $d_1 s_0 = \id$, it follows that $d_0 y = d_1 y = x$. Furthermore $d_i y = d_i s_0 x = s_0 d_{i-1} x = 0$ for all $i > 1$, because $x \in Z_n(X)$.
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\emph{Symmetry}. Let $x, x' \in Z_n(X)$ with $y: x \sim x'$. Define $y' = s_0 x + s_0 x' - y$, then by using linearity: $d_0 y' = x + x' - x = x'$ and $d_1 y' = x + x' - x' = x$. For $i>1$ we again get $d_i y' = 0$, because $x \in Z_n(X)$.
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\emph{Transitivity}. Let $x_0, x_1, x_2 \in Z_n(X)$ with $y: x_0 \sim x_1$ and $z: x_1 \sim x_2$. Define $w = y + z - s_0 x_1$. By linearity we have $d_0 w = x_0 + x_1 -x_1 = x_0$, similarly $d_1 w = x_2$. Again for $i>1$ we have $d_i w = 0$.
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\emph{Addition}. Let $y: x_0 \sim x_1$ and $z: x_2 \sim x_3$. Then by linearity $y + g: x_0 + x_2 \sim x_1 + x_3$ and $-y: -x_0 \sim -x_1$.
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\end{proof}
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\begin{definition}
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Given a simplicial abelian group $X$, we define the $n$-th homotopy group as:
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$$ \pi_n(X) = Z_n(X) / \sim. $$
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\end{definition}
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Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and $\sim$ also defines a subgroup. It is relatively straight forward to prove that this definition coincides with the $n$-th homology group of the associated normalized chain complex.
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\begin{lemma}
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For any simplicial abelian group $X$:
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$$ \pi_n(X) = H_n(N(X)). $$
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\end{lemma}
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\begin{proof}
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By writing out the definitions of the $n$-cycles and $n$-boundaries of the normalized chain complex, we see:
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\begin{align*}
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\ker(\del) &= \{ x \in N(X)_n \I \del(x) = 0 \} \\
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&= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i > 0 \text{ and } d_0(x) = 0 \} \\
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&= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i \leq n \} \\
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&= Z_n(X) \\
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\im(\del) &= \{ \del(y) \I y \in N(X)_{n+1} \} \\
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&= \{ d_0 y \I y \in X_{n+1}, d_i(y) = 0 \text{ for all } i > 0 \} \\
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&= \{ x \in N(X)_n \I x \sim 0 \}
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\end{align*}
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So we see that $\pi_n(X) = Z_n(X) / \sim = \ker(\del) / \im(\del) = H_n(N(X))$.
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\end{proof}
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\begin{corollary}
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For a chain complex $C$ we have $H_n(C) \iso \pi_n(K(C))$
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\end{corollary}
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\begin{proof}
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By the established equivalence we have for any chain complex $C$:
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$$ \pi_n(K(C)) \iso H_n(N(K(C))) \iso H_n(C). $$
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\end{proof}
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\subsection{Topology}
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In section~\ref{sec:Constructions} we saw that we can construct a functor $G: \cat{C} \to \sSet$ if we are provided a functor the other way around. If we can define a functor $F: \DELTA \to \Top$, then for any space $X$ we have a simplicial set $\Hom{\Top}{F-}{X}: \DELTA^{op} \to \Set$. In section~\ref{sec:Chain Complexes}, we already defined the \emph{topological $n$-simplex} $\Delta^n$ and face maps $\delta^i : \Delta^n \mono \Delta^{n+1}$. We can similarly define degeneracy maps $s^i: \Delta^n \to \Delta^{n-1}$ as:
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$$ s^i(x_0, \ldots, x_n) = (x_0, \ldots, x_i + x_{i+1}, \ldots, x_n) \in \Delta^{n-1}. $$
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The reader is invited to check the cosimplicial identities himself and conclude that we now have a functor $F: \DELTA \to \Top$, and hence we have a functor $S: \Top \to \sSet$ given by:
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$$ \text{Sing}(X)_n = \Hom{\Top}{\Delta^n}{X}. $$
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Recall construction of the singular chain complex in section~\ref{sec:Chain Complexes}:
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$$ C_n(X) = \Z[\Hom{\cat{Top}}{\Delta^n}{X}]. $$
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Where the boundary map was given as an alternating sum. Looking more closely we see that this construction decomposes as:
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$$ C: \Top \tot{\text{Sing}} \sSet \tot{\Z^\ast} \sAb \tot{C} \Ch{\Ab}, $$
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where the last functor is the \emph{unnormalized chain complex}. All the categories involved have a notion of homotopy. In topological spaces this is the known notion where $f, g:X \to Y$ are homotopic if there exists a homotopy $H:I \times X \to Y$ with the appropriate properties. In simplicial sets (or simplicial abelian groups) we only saw the notion of homotopy groups, but there exists a more general notion of homotopy, as discussed in the overview of Friedman \cite{friedman}. And finally in chain complexes we saw homology groups, but this category also has a more general notion of chain homotopy, which can be found in any book on homological algebra such as in the book of Weibel \cite{weibel}.
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It is known that for any simplicial abelian group both the normalized and unnormalized chain complex have the same homology groups. More precisely for any simplicial abelian group $X$ we have:
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$$ H_n(N(X)) \iso H_n(C(X)) \quad\text{for all } n \in \N. $$
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This is for example proven in \cite[Theorem 4.1]{eilenberg}. So this assures that the homology groups of the singular chain complex of a space are really the homotopy groups of the simplicial abelian group which is in the background.
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