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I'm working thru Hatcher on my own and am having trouble with basic definitions relating to his Cellular Boundary Formula on p 140.

What is the domain and range of the function in question? Do the d(.)expressions refer to the composite boundary maps he defined on the previous page? If so, how can a map be defined as a sum of other maps? Or, do the d(.) terms refer to degrees? If so, what does the degree of an n-cell mean?

Obviously, I need some help here. Thanks.

PossumP
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  • I'm not sure I understand the source of your confusion. As explained in the sentence following the formula, the $d_{\alpha \beta}$ are just the degrees of appropriate maps (the compositions $S_\alpha^{n-1} \to X^{n-1} \to S^{n-1}\beta)$. The map $d_n$ goes from $H_n(X^n,X^{n-1})$ (generated by $n$-cells) to $H{n-1}(X^{n-1},X^{n-2})$ (generated by $(n-1)$-cells). – Alex Provost Apr 13 '17 at 23:26
  • This must be a dumb question, but it seems like d on the left hand side of the equation is a map, but the d terms on the r.h.s. are integers. That doesn't make sense. – PossumP Apr 13 '17 at 23:59
  • The $d_n(e^n_\alpha)$ on the left-hand side is the result of $d_n$ applied to $e^n_\alpha$ (viewed as a generator of $H_n(X^n,X^{n-1})$), so it should yield an element of $H_{n-1}(X^{n-1},X^{n-2})$. The sum on the right-hand side is an integral linear combination of $(n-1)$-cells, viewed as generators of $H_{n-1}(X^{n-1},X^{n-2})$. So it does make sense. – Alex Provost Apr 14 '17 at 00:06
  • And yes, $d_n$ and $d_{\alpha \beta}$ refer to totally different objects. The author is not claiming equality between a $d_n$ and a sum of $d_{\alpha \beta}$s. – Alex Provost Apr 14 '17 at 15:16

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The author is not claiming equality between a $d_n$ and a sum of $d_{\alpha \beta}$s. These are different types of objects, as you've noted. Rather, what is true is that the result of $d_n$ applied to the $n$-cell $e^n_\alpha$ (viewed as a generator of $H_n(X^n,X^{n-1})$) can be computed as the sum of all the $(n-1)$-cells $e^{n-1}_\beta$ (viewed as generators of $H_n(X^{n-1},X^{n-2})$), each appearing exactly $d_{\alpha \beta}$ times.

To be very explicit (and this is just linear algebra): one of the first things that we learn about the groups $C_n:=H_n(X^n,X^{n-1})$ defining the cellular chain complex is that they are isomorphic to the free Abelian groups $\mathbb{Z}\langle e^n_\alpha \rangle$ generated by the $n$-cells of the complex. Call this isomorphism $$\phi:C_n \xrightarrow{\sim} \mathbb{Z}\langle e^n_\alpha \rangle.$$

Then, as we often do in mathematics, we will treat the domain and the codomain of $\phi$ as the same object.

Under this correspondence, the data of a homomorphism $f:C_n \to A$ to another Abelian group $A$ is equivalent to the data of a homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to A$ by the relation $f = g\circ \phi$. But $\mathbb{Z}\langle e^n_\alpha \rangle$ is free Abelian with basis the $n$-cells of the complex, so any homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to A$ is entirely determined by its action on each basis element (each $n$-cell). This is because if you know $g(e^n_\alpha)$ for every $\alpha$, you know everything:$$g(\sum_\alpha c_\alpha e^n_\alpha) = \sum_\alpha c_\alpha g(e^n_\alpha),$$ where each $c_\alpha \in\mathbb{Z}$ is a scalar.

Similarly, the data of a homomorphism $f:C_n \to C_{n-1}$ is equivalent to the data of a homomorphism $g:\mathbb{Z}\langle e^n_\alpha \rangle \to \mathbb{Z}\langle e^{n-1}_\beta \rangle$ by the conjugation relation $f = \phi^{-1} \circ g\circ \phi$.

So, formally, the claim is that $d_n = \phi^{-1} \circ g \circ \phi$, where $g$ is defined on the basis elements by $$g(e^n_\alpha) = \sum_{\beta}d_{\alpha \beta}e^{n-1}_\beta.$$

Alex Provost
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  • If I understand you correctly, you are saying Hatcher is being somewhat loose with his notation (in his use of the = sign, and in defining the two d's differently). How, then, can the cellular boundary formula be written in completely clear mathematical terms (without a lot of "interpretive" comments)? – PossumP Apr 18 '17 at 23:38
  • @MPitts No, he is not being loose at all. I really don't see the issue here; would you object to using the symbols $A$ for a matrix and $(A_{i,j})$ for its components? There is no room for confusion, given that one letter has no indices and the other has none. Then we can write $A(e_j) = \sum_i A_{i,j} f_i$ for the chosen bases $(e_j),(f_i)$ of the domain and codomain. It's exactly the same thing here: $d_n$ is the linear map, $(e^n_\alpha),(e^{n-1}\beta)$ are the bases, and the transpose of $(d{\alpha \beta})$ is the matrix of $d_n$ with respect to these bases. – Alex Provost Apr 19 '17 at 12:59
  • @MPitts I meant to write "There is no room for confusion, given that one letter has two indices and the other has none." – Alex Provost Apr 19 '17 at 14:21
  • Ok, then for the map d (on the l.h.s.) the domain and codomain are bases (not cells or groups, or anything else). Right? – PossumP Apr 19 '17 at 16:59
  • @MPitts Have a look at my edit, where I (hopefully) laid things out in great detail. – Alex Provost Apr 19 '17 at 17:58
  • Got it. I REALLY appreciate all your explanations - I was having a mental block that I just could not surmount. – PossumP Apr 20 '17 at 21:47
  • @MPitts Glad it worked out :-) I know exactly how that feels! – Alex Provost Apr 20 '17 at 22:06