# Hatcher's formula in homotopy equivalence proof

In the proof that two homotopic maps induce the same homomorphism in homology, appears the formula (bottom of p. 112, Hatcher, Algebraic Topology): \begin{gather} P(\partial \sigma) = \sum_{i<j} (-1)^i(-1)^j F \circ (\sigma \times id_I) |[v_0 \dots v_i, w_i \dots \hat w_j \dots w_n] + \\ \sum_{i>j} (-1)^{i-1}(-1)^j F \circ (\sigma \times id_I) |[v_0 \dots \hat v_j \dots v_i, w_i \dots \dots w_n]. \end{gather}

Why is it so? I really can't see how this is computed.

Context: the prism operator $P : C_n(X) \rightarrow C_{n+1}(Y)$ is defined as $$P(\sigma) := \sum_i (-1)^i F \circ (\sigma \times id_I) | [v_0 \dots v_i, w_i \dots w_n] ,$$ $F$ is the homotopy, $I := [0,1]$, $[v_0 \dots v_i, w_i \dots w_n]$ is a $n$-simplex of the subdivision into simplexes of $\Delta_n \times I$.

• $[v_0 \dots v_i, w_i \dots w_n]$ is an $(n+1)$-simplex – user153312 Jan 5 '15 at 12:04

First of all once computes $\partial \sigma$ where $\sigma$ is understood to be $\sigma : [v_0,v_1,.....,v_n] \rightarrow X$.

$\partial \sigma = \sum_i (-1)^i\sigma|[v_0,....,v_{i-1}, \hat{v}_i,...v_n]$.

Now use the formula for the prism operator. Now we have to be careful with the terms. Here we are just applying the prism operator to each of the $n-1$ simplex in the boundary. so calculating we have

$P(\sigma|[v_0,....,v_{i-1}, \hat{v}_i,...v_n]) = \sum_{i>j} (-1)^jF\circ (\sigma \times Id)| [v_0,....,v_j,w_j,.... \hat{w}_i,...w_n] + \sum_{i<j} (-1)^{j-1}F\circ (\sigma \times Id)|[v_0,...,\hat{v}_i,...,v_j,w_j,...,w_n]$.

Now when we sum this over $i$ with the sign $(-1)^i$, on the LHS we get $P(\partial \sigma)$ and on the RHS the formula mentioned.

• Thank you, but why it is $P([v_0 \dots \hat v_i \dots v_n])$ on the LHS and not $P(\sigma|[v_0 \dots \hat v_i \dots v_n])$? Because the problem is I get $$P(\sigma|[v_0 \dots \hat v_i \dots v_n]) = \sum (-1)^j F \circ (\sigma|[v_0 \dots \hat v_i \dots v_n] \times id) | [v_0 \dots v_j,w_j \dots w_{n-1}]$$ and then it's a mistery... – Aldebaran Sep 21 '14 at 7:28
• Yes. Thank you. That was a typo. – random123 Sep 21 '14 at 7:30
• But how do you get from $$P(\sigma|[v_0 \dots \hat v_i \dots v_n]) = \sum (-1)^j F \circ (\sigma|[v_0 \dots \hat v_i \dots v_n] \times id) | [v_0 \dots v_j,w_j \dots w_{n-1}]$$ to the RHS of your equation? – Aldebaran Sep 21 '14 at 7:34
• There is another term in addition to the term you wrote in the expression for $P(\sigma|[v_0,.......\hat{v}_i,...,vn])$. The term with $i<j$. – random123 Sep 21 '14 at 7:39