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First, this is the link to the book, for convenience: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf#page=275

Corollary 3A.7.

(a) $ \widetilde{H}_n (X;\Bbb Z)=0$ for all $n$ iff $\widetilde{H}_n(X;\Bbb Q)=0$ and $\widetilde{H}_n(X;\Bbb Z_p)=0$ for all $n$ and all primes $p$.

(b) A map $f:X \to Y$ induces isomorphisms on homology with $\Bbb Z$ coefficients iff it induces isomorphisms on homology with $\Bbb Q$ and $\Bbb Z_p$ coefficients for all primes $p$.

I understood the proof of (a), but not for (b). The proof of (b) is as follows:

Proof. Statement (b) follows from (a) by passing to the mapping cone of $f$.

I have really no idea for this proof.

In fact, this question is related to the question linked below: Using Mapping cone to show map induce isomorphism on homology

However, in the above link, I can't see

  1. how the long exact sequence including the mapping cone is derived. (Is the sequence valid for any coefficients?)

  2. why the reduced homology groups of the mapping cone are zero. (Is this because the map $f$ induces isomorphisms on homology?)

  3. how the sequence implies the result.

So I made another question. Thanks in advance.

William
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blancket
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1 Answers1

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$1)$ For a continuous function $f\colon X \to Y$ the long exact sequence for the mapping cone comes from the long exact sequence for the pair $(M_f, X)$, where the mapping cylinder $M_f$ is homotopy equivalent to $Y$ and $C_f = M_f / X$. Since $(M_f, X)$ is a "good" pair the quotient map induces an isomorphism of homology groups $H_n(M_f, X; G) \cong \tilde{H}_n(C_f; G)$ for any $G$.

$2)$ If $f$ induces isomorphisms $H_n(X;G)\cong H_n(Y;G)$ for each $n$, then by exactness of the mapping cone sequence $\tilde{H}_n(C_f;G) \cong 0$ for all $n$. In fact the converse is also true: if $\tilde{H}_n(C_f;G) \cong 0$ for all $n$ then $f$ is an $H_*(-;G)$ isomorphism.

With these facts you should be able to use part $(a)$ to deduce part $(b)$.

William
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