# Excisive triads and weak equivalences

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).

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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

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.