Understanding the proof of a theorem in Hatcher

Question

I don't understand the proof of theorem 2.10 in the book (page 112) of Allen Hatcher on algebraic topology.

It states: Two homotopic maps induce the same homomorphism on homology groups.

Here is the part I'm stuck with:

When he calculates $\partial P(\sigma)$, I do not understand how the terms with $i = j$ cancel, and how we are left with the terms he says we do. I tried to do this explicitly for an easy example but it didn't work out. Any help is appreciated here.

Also, I don't understand why the terms with $i \neq j$ are $- P \partial (\sigma)$.

Answer 1

Although Hatcher's result is correct, his "cancellation" explanation needs clarification.

For a singular $m$-simplex $\tau : \Delta^m \to X$, where $\Delta^m = [u_0,\dots,u_m]$, we write $d_k(\tau) = \tau \mid [u_0,\dots,\hat{u}_k,\ldots,u_m]$ for $k = 0,\ldots, m$.

Let us abbreviate $F \circ (\sigma \times \mathbb I)$ by $\phi$. Then we have $$P(\sigma) = \sum_{i=0}^n (-1)^i \phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]$$ $$\partial P(\sigma) = \sum_{i=0}^n (-1)^i \partial (\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{i=0}^n (-1)^i \sum_{j=0}^{n+1} (-1)^jd_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{i,j} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\ = \sum_{j \le i \le n} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) + \\ \sum_{n+1 \ge j > i} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{j \le i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] + \\ \sum_{n+1 \ge j > i} (-1)^{i+j}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j-1},\ldots,w_n] \\ = \sum_{j \le i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] + \\ \sum_{n \ge j \ge i} (-1)^{i+j+1}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j},\ldots,w_n]$$ Now you see that each term indexed by $(i,i)$ with $n \ge i > 0$ in the first sum cancels with the term indexed by $(i-1,i-1)$ in the second sum. Thus only the terms indexed by $(0,0)$ in the first sum and by $(n,n)$ in the second sum survive. These are $\phi \mid [w_0,\ldots,w_n] = g_\#(\sigma)$ and $-\phi \mid [v_0,\ldots,v_n] = -f_\#(\sigma)$. Thus we get $$\partial P(\sigma) - g_\#(\sigma) + f_\#(\sigma) = \\\sum_{j < i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] +\\ \sum_{n \ge j > i} (-1)^{i+j+1}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j},\ldots,w_n]$$ For each face $d_j(\sigma) = \sigma \mid [v_0,\ldots,\hat v_j,\ldots, v_n]$ we have to consider the prism with bottom $[v_0,\ldots,\hat v_j,\ldots, v_n]$ and top $[w_0,\ldots,\hat w_j,\ldots, w_n]$. We get $$P \partial \sigma = P(\sum_{j=0}^n (-1)^j d_j(\sigma)) = \sum_{j=0}^n (-1)^j P(d_j(\sigma)) = \sum_{j=0}^n (-1)^j P(\sigma \mid [v_0,\ldots,\hat v_j,\ldots, v_n])=\\ \sum_j (-1)^j \left(\sum_{i<j}(-1)^i \phi \mid [v_0,\ldots,v_i,w_i,\ldots\hat w_j,\ldots, w_n] + \\ \sum_{i\ge j}(-1)^i \phi \mid [v_0,\ldots,\hat v_j,\ldots, v_{i+1},w_{i+1},\ldots w_n]\right) \\= \sum_{i < j} (-1)^{i+j} \phi \mid [v_0,\ldots,v_i,w_i,\ldots\hat w_j,\ldots, w_n] + \\ \sum_{i > j} (-1)^{i+j+1} \phi \mid [v_0,\ldots,\hat v_j,\ldots, v_{i},w_{i},\ldots w_n]$$

Answer 2

For the first part, the idea is that the term from the first sum with $i=j$ will cancel from the term in the second sum with $i-1=j-1$. It might be better to write the first sum as $\sum_{i=1}^{n} \sum_{j=1}^{i} ...$ and similarly for the second sum. The $(i,i)$ term in the first sum will cancel with the $(i-1,i-1)$ term in the second sum: the signs will be opposite, and the simplices from the two sums will be the same: $$ [v_0, ..., v_{i-1}, \hat{v}_i, w_i, ..., w_n], \quad [v_0, ..., v_{i-1}, \hat{w}_{i-1}, w_i, ..., w_n]. $$ The remaining terms will come from the first sum when $i=0$ and from the second sum when $i=n$, since those are not part of the cancellation.

For the second part ($i \neq j$ terms), what part of Hatcher's explanation is not clear?