equal
deleted
inserted
replaced
33 \begin{proof} |
33 \begin{proof} |
34 (Sketch) |
34 (Sketch) |
35 This is a standard result; see, for example, \nn{need citations}. |
35 This is a standard result; see, for example, \nn{need citations}. |
36 |
36 |
37 We will build a chain map $f\in \Compat(D^\bullet_*)$ inductively. |
37 We will build a chain map $f\in \Compat(D^\bullet_*)$ inductively. |
38 Choose $f(x_{0j})\in D^{0j}_0$ for all $j$. |
38 Choose $f(x_{0j})\in D^{0j}_0$ for all $j$ |
39 \nn{...} |
39 (possible since $D^{0j}_0$ is non-empty). |
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40 Choose $f(x_{1j})\in D^{1j}_1$ such that $\bd f(x_{1j}) = f(\bd x_{1j})$ |
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41 (possible since $D^{0l}_* \sub D^{1j}_*$ for each $x_{0l}$ in $\bd x_{1j}$ |
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42 and $D^{1j}_*$ is 0-acyclic). |
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43 Continue in this way, choosing $f(x_{kj})\in D^{kj}_k$ such that $\bd f(x_{kj}) = f(\bd x_{kj})$ |
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44 We have now constructed $f\in \Compat(D^\bullet_*)$, proving the first claim of the theorem. |
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45 |
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46 Now suppose that $D^{kj}_*$ is $k$-acyclic for all $k$ and $j$. |
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47 Let $f$ and $f'$ be two chain maps (0-chains) in $\Compat(D^\bullet_*)$. |
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48 Using a technique similar to above we can construct a homotopy (1-chain) in $\Compat(D^\bullet_*)$ |
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49 between $f$ and $f'$. |
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50 Thus $\Compat(D^\bullet_*)$ is 0-connected. |
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51 Similarly, if $D^{kj}_*$ is $(k{+}i)$-acyclic then we can show that $\Compat(D^\bullet_*)$ is $i$-connected. |
40 \end{proof} |
52 \end{proof} |
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53 |
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54 \nn{do we also need some version of ``backwards" acyclic models? probably} |