Joshua Moerman
12 years ago
4 changed files with 69 additions and 3 deletions
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\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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\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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$$ Z_n(X) = \{ x \in X_n | d_i(x) = \ast \text{ for all } i < n \}. $$ |
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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_{n+1}(y) &= x' \\ |
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d_n(y) &= x \\ |
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d_i(y) &= \ast \text{ for all } i < n. |
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\end{align} |
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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: Insert prove $\sim$ is eq. rel. for $\sAb$.} |
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\todo{Htp: Discuss/picturize Kan-condition?} |
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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 < n \text{ and } d_n(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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\end{align*} |
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\begin{align*} |
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\im(\del) &= \{ \del(y) \I y \in N(X)_{n+1} \} \\ |
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&= \{ d_{n+1} \I y \in X_{n+1}, d_i(y) = 0 \text{ for all } i \leq n \} \\ |
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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: |
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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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