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In the proof of barycentric subdivision in singular homology, we take the subdivision operator $b: C_q(X) \to C_q(X)$, and do some algebra to define an operator $b^\infty$, which applies $b$ "as many times as needed".

Consider the following attempt to express the lemma algebraically (from Ramanan's Global Calculus):

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Implicit here is the claim that $N(x) \geq N(y_j)$ for all $j$ (Edit: and, in fact, we would also need that $N(y_j) \geq N(ky_j)$, as Eric points out). This is indeed true in the geometric case where we use it--since if $x$ is a simplex, and $b^{N(x)}(x)$ is in some $U$ in a cover, then clearly $b^{N(x)}(y_j)$ is as well. But in the more general setting of the above lemma, I'm not convinced that $N(x) \geq N(y_j)$ is necessarily true.

We can do a little better if we define $k_n$ the reverse way: $k_n:= k \circ (1 + b + \dotsb + b^{n-1})$. This still works as a homotopy $\mathrm{Id} \Rightarrow b^n$, and on the $y_j$ we end up with $(k_\infty - k_{N(x)})(y_j) = k\left(\sum\limits_{i=N(y_j)}^{N(x)-1}b^i(y_j)\right)$. But we still need $N(x) \geq N(y_j)$.

As it is, the proof seems unsalvagable without more hypotheses. Do you agree?

Eric Auld
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  • Am I missing something, or don't you also need to know $N(y_j)\geq N(ky_j)$, since you need $b^iky_j$ to be in $L$, not $b^iy_j$? – Eric Wofsey May 22 '16 at 19:25
  • @EricWofsey You're right! You say "also", but I think that's all we need, right? – Eric Auld May 22 '16 at 21:04
  • Well, you also need $N(x)\geq N(ky_j)$: you need to know that $b^iky_j$ is in $L$ whenever $i$ is between $N(y_j)$ and $N(x)$, so you need $\min(N(x),N(y_j))$ to be $\geq N(ky_j)$. – Eric Wofsey May 22 '16 at 21:09
  • @EricWofsey We could define $k_n$ the other way, $k_n:=k \circ(1 + \dotsb + b^{n-1})$, and that allows us to assume only $N(x) \geq N(y_j)$, I believe. – Eric Auld May 22 '16 at 22:58

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