equal
deleted
inserted
replaced
13 \end{lem} |
13 \end{lem} |
14 \begin{rem} |
14 \begin{rem} |
15 We can't quite do the same with all $\cV_k$ just equal to $\cU$, but we can get by if we give ourselves arbitrarily little room to maneuver, by making the blobs we act on slightly smaller. |
15 We can't quite do the same with all $\cV_k$ just equal to $\cU$, but we can get by if we give ourselves arbitrarily little room to maneuver, by making the blobs we act on slightly smaller. |
16 \end{rem} |
16 \end{rem} |
17 \begin{proof} |
17 \begin{proof} |
|
18 This follows from the remark \nn{number it and cite it?} following the proof of |
|
19 Proposition \ref{CHprop}. |
|
20 \end{proof} |
|
21 \noop{ |
18 We choose yet another open cover, $\cW$, which so fine that the union (disjoint or not) of any one open set $V \in \cV$ with $k$ open sets $W_i \in \cW$ is contained in a disjoint union of open sets of $\cU$. |
22 We choose yet another open cover, $\cW$, which so fine that the union (disjoint or not) of any one open set $V \in \cV$ with $k$ open sets $W_i \in \cW$ is contained in a disjoint union of open sets of $\cU$. |
19 Now, in the proof of Proposition \ref{CHprop} |
23 Now, in the proof of Proposition \ref{CHprop} |
20 \todo{I think I need to understand better that proof before I can write this!} |
24 [...] |
21 \end{proof} |
25 } |
22 |
26 |
23 |
27 |
24 \begin{proof}[Proof of Theorem \ref{thm:small-blobs}] |
28 \begin{proof}[Proof of Theorem \ref{thm:small-blobs}] |
25 We begin by describing the homotopy inverse in small degrees, to illustrate the general technique. |
29 We begin by describing the homotopy inverse in small degrees, to illustrate the general technique. |
26 We will construct a chain map $s: \bc_*(M) \to \bc^{\cU}_*(M)$ and a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ so that $\bdy h+h \bdy=i\circ s - \id$. The composition $s \circ i$ will just be the identity. |
30 We will construct a chain map $s: \bc_*(M) \to \bc^{\cU}_*(M)$ and a homotopy $h:\bc_*(M) \to \bc_{*+1}(M)$ so that $\bdy h+h \bdy=i\circ s - \id$. The composition $s \circ i$ will just be the identity. |