I'm trying to follow an argument in Lück's "Algebraische Topologie: Homologie und Mannigfaltigkeiten" (to which there apparently doesn't exist an english translation). The aim is to check homotopy invariance of cellular homology by constructing a chain homotopy.

Let me sketch the argument. Let $h\colon (X,A)\times [0,1]\to (Y,B)$ be a cellular homotopy from $f_0$ to $f_1$. Using the CW-structure on $[0,1]$ with the two 0-cells {0}, {1} and one 1-cell, we identify $$ C_n((X,A)\times [0,1])=C_n(X,A)\oplus C_n(X,A)\oplus C_{n-1}(X,A). $$ Then $C_n(h)$ is of the form $C_n(f_0)\oplus C_n(f_1)\oplus u_{n-1}$, where $u_{n-1}$ is some map $C_{n-1}(X,A)\to C_n(Y,B)$.

Now we would like to compute the $n$th differential of $C_*((X,A)\times [0,1])$ under the above identification. Since it is a map from $C_n(X,A)\oplus C_n(X,A)\oplus C_{n-1}(X,A)$ to $C_{n-1}(X,A)\oplus C_{n-1}(X,A)\oplus C_{n-2}(X,A)$, we can denote it by a 3x3-matrix. My computation yielded $$ \begin{pmatrix} c_n & 0 & 0 \newline 0 & c_n & 0 \newline -id & id & c_{n-1} \end{pmatrix}, $$ where $c_n$ is the $n$th differential of $C_*(X,A)$. In the book, however, I find $$ \begin{pmatrix} c_n & 0 & (-1)^n \newline 0 & c_n & (-1)^{n-1} \newline -id & id & c_{n-1} \end{pmatrix}. $$

Question: What is meant by $(-1)^n\colon C_{n}(X,A)\to C_{n-2}(X,A)$ and how can I understand that the second matrix above is the correct expression?

Edit: Apparently, the correct form of the the matrix representing the $n$th differential of $C_*((X,A)\times [0,1])$ is $$ \begin{pmatrix} c_n & 0 & 0 \newline 0 & c_n & 0 \newline (-1)^{n+1}\cdot id & (-1)^n\cdot id & c_{n-1} \end{pmatrix} $$ or $$ \begin{pmatrix} c_n & 0 & 0 \newline 0 & c_n & 0 \newline (-1)^{n}\cdot id & (-1)^{n+1}\cdot id & c_{n-1} \end{pmatrix}, $$ depending on the orientation of the 1-cell in $[0,1]$.

  • $\begingroup$ Isn't there a straightforward conceptual way to prove this from the proof for singular homology by proving that simplicial sets are Quillen equivalent to CG(WH)-spaces? $\endgroup$ – Harry Gindi Jul 11 '10 at 16:40
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    $\begingroup$ That's not particularly relevant to the OP's question, Harry. This is about giving a proof in a specific context of interest, not translating it to another context where you give an equivalent proof. $\endgroup$ – Ryan Budney Jul 12 '10 at 2:19

Your calculation is almost correct. I don't have a copy of the book you are referring to, but if it says what you claim it does, then this is a typo. This clearly must be, since the matrix in question is supposed to be a differential yet only squares to zero if the mysterious map "(-1)^n" is zero.

However your matrix also suffers from a similar problem. It doesn't square to zero either. The problem seems to be about orienting the cells correctly. You'll find that the signs on the "id" terms of your matrix need to alternate at each level.

  • $\begingroup$ I believe you are right. Thank you. I edited the question accordingly. This way, I also obtain the equation showing that $v_n=(-1)^n\cdot u_n$ is a chain homotopy between $C_n(f_0)$ and $C_n(f_1)$. $\endgroup$ – Rasmus Jul 12 '10 at 13:23

I hope it will be interesting for topologists to see this result on cellular homotopy as a special case of a result on filtered spaces $X_\ast$ and of the homotopy invariance of the fundamental crossed complex functor

$\Pi: ($filtered spaces$)\to ($crossed complexes$)$

defined using the fundamental groupoid and relative homotopy groups. The difficult part is to construct a natural transformation

$\zeta : \Pi X_\ast \otimes \Pi Y_\ast \to \Pi (X_\ast \otimes Y_\ast )$

(which is an isomorphism for cellular filtrations) and this uses cubical methods. This is all explained in our book published by the EMS and advertised here.

  • $\begingroup$ Dear Ronnie Brown, thanks for sharing your viewpoint on this. It is always interesting to hear what advanced people consider a given elementary object or fact as a special case of. $\endgroup$ – Rasmus Aug 12 '12 at 14:39
  • $\begingroup$ @Rasmus: I wondered, mused?, if your comment was ironic!? The book shows a new foundation for algebraic topology, particularly the relation between homology and homotopy, without using Poincar\'es device of "formal sums", but instead actual compositions for elements of homotopically defined functors. $\endgroup$ – Ronnie Brown Dec 15 '13 at 11:37
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    $\begingroup$ No, my comment was not ironic. I meant what I wrote. $\endgroup$ – Rasmus Dec 15 '13 at 11:42
  • $\begingroup$ @Rasmus: Great! Actually the argument for the book was to put all this material in one place to make it easier to evaluate. As Einstein wrote:"Concepts which have proved useful for ordering things easily assume so great an authority over us, that we forget their terrestrial origin and accept them as unalterable facts." There are some anomalies at the beginnings of algebraic topology (see my Chicago seminar (2012) on my preprint page) and one needs to compare and contrast approaches. $\endgroup$ – Ronnie Brown Dec 16 '13 at 22:02

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