2

Let us suppose that $f(X',A',B',x_0') \rightarrow (X;A,B,x_0)$ is a map of triads such that $$f_\ast:\pi_\ast (A' \cap B',x_0') \rightarrow \pi_\ast(A \cap B, x_0)$$ $$f_\ast:\pi_\ast(A',x_0')\rightarrow \pi_\ast(A,x_0)$$ and $$f_\ast : \pi_(B',x_0') \rightarrow \pi_\ast(B,x_0)$$ are all isomorphisms. Show that if X is excisive (meaning, it is the union of the interiors of A and B), then $$f_\ast : \pi(X';A',B',x_0') \rightarrow \pi_(X;,A,B,x_0)$$ is an isomorphism.

I have been trying to follow the hint that is given, namely that we should replace X be a mapping cylinder, but I can't seem to find an appropriate one to work with (that is excisive for example).

Tedar
  • 529
  • Probably both triads must be excisive. Let $X' = X \vee S^2$, $A'$ and $B'$ like in $X$, ignoring the sphere. All conditions are trivially true (they do not care about anything not in $A'$ and $B'$) but the triads are not homotopy equivalent. – Dmitry Aug 25 '13 at 11:18
  • If I'm not mistaken, by several applications of five-lemma (to exact sequences of pairs and exact sequence of a triad, binding $\pi_k(X; A, B, )$ to pairs $\pi_k(X; B, )$ and $\pi_k(A; A \cap B, *)$) the statement is equivalent to $X' \to X$ being a weak equivalence. This seems unlikely, though I haven't thought about it long enough. – Dmitry Aug 25 '13 at 11:23

1 Answers1

2

There is a proof of equivalent statement ($X' \to X$ is a weak equivalence) in Peter May's book, page 80, the last theorem in the section "Approximation of excisive triads by CW triads". It's not particularly short and appeals to some lemmas from previous sections. Unfortunately proofs from this book tend to be somewhat formal.

Amusingly, I've first made up a proof for excisive CW-triads (i.e. $A$ and $B$ are not subsets covering $X$ by their interiors but subcomplexes of $X$ such that every cell of $X$ is contained in one of them) and consulted the book only to check whether the argument by CW-approximation wasn't circular and indeed it was.

Dmitry
  • 1,238
  • Thank you! I knew of this proof actually, but I thought that maybe the hint that was given (namely, to replace X by a mapping cylinder) would give some easier proof, but I haven't found one so far. Tell me if you can find one - I'll wait a few days with accepting the answer nonetheless, to see if something else comes up. Thanks you for your help! – Tedar Aug 25 '13 at 13:50
  • By the way, if you find a simpler proof yourself, post it here as an answer. Answering one's own question is OK on this site. Good luck! – Dmitry Aug 25 '13 at 14:37