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Let $f :X \to Y$ be a continuous map of spaces show the following conditions are equivalent

  1. $f_* : H_n( X) → H_n(Y)$ is an isomorphism for all $n \geq 0$

  2. $f^* :H^n(Y,\mathbb{Z}) \to H^n(X,\mathbb{Z})$ is an isomorphism for all $n \geq 0$.

I think I need to use universal coefficient theorem and the theorem below

if $G$ is an abelian group such that $\text{Hom}(G,Z)=0$ and $\text{Ext}(G,Z)=0$ then $G=0$.

But I cannot write down the details.

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By naturality of the universal coefficient theorem for cohomology, $f$ induces a ladder diagram that looks like

$$\require{AMScd} \begin{CD} 0 @>>> \operatorname{Ext}(H_{n-1}(X),\mathbb{Z}) @>>> H^n(X,\mathbb{Z}) @>>> \operatorname{Hom}(H_n(X),\mathbb{Z}) @>>> 0\\ @. @A{\cong}AA @AAA @A{\cong}AA @.\\ 0 @>>> \operatorname{Ext}(H_{n-1}(Y),\mathbb{Z}) @>>> H^n(Y,\mathbb{Z}) @>>> \operatorname{Hom}(H_n(Y),\mathbb{Z}) @>>> 0\\ \end{CD}$$

So by the (short) five lemma, the middle arrow is an isomorphism when $f_*:H_n(X) \to H_n(Y)$ is.

For the converse, see Isomorphism on Cohomology implies isomorphism on homology.

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  • $\begingroup$ I can understand your answer, but the link you give about converse, I still cannot understand. $\endgroup$ – noname1014 May 15 '17 at 4:48
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Note that $(1)$ is equivalent to $\tilde H_*(M_f)=0$ and $(2)$ is equivalent to $\tilde H^*(M_f)=0$ where $M_f$ is the mapping cylinder of $f$.

Now use universal coefficients and your theorem to conclude the implication $(2)\implies (1)$.

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