1
$\begingroup$

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 $\Delta_n \times I$.

$\endgroup$
  • $\begingroup$ $[v_0 \dots v_i, w_i \dots w_n]$ is an $(n+1)$-simplex $\endgroup$ – user153312 Jan 5 '15 at 12:04
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – Aldebaran Sep 21 '14 at 7:28
  • $\begingroup$ Yes. Thank you. That was a typo. $\endgroup$ – random123 Sep 21 '14 at 7:30
  • $\begingroup$ 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? $\endgroup$ – Aldebaran Sep 21 '14 at 7:34
  • $\begingroup$ 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 $i<j$. $\endgroup$ – random123 Sep 21 '14 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.