Injection on homology dual to surjection on cohomology?

Question

Let $f:X\to Y$ be morphism between two topological spaces. We know that

$H_k(X)\xrightarrow{f_*} H_k(Y)$ is injective if $H^k(Y)\xrightarrow{f^*} H^k(X)$ is surjective.

I want to know if the converse also true? namely,

$H_k(X)\xrightarrow{f_*} H_k(Y)$ is injective only if $H^k(Y)\xrightarrow{f^*} H^k(X)$ is surjective.

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My approach

The "if" part is easy: if $f^*$ surjective, then for any $a\neq b$ in $H_k(X)$, we have $$<a,f^*\alpha>\neq<b,f^*\alpha>$$ for some $\alpha$, which is just $$<f_*a,\alpha>\neq<f_*b,\alpha>$$ hence $f_*a\neq f_*b$.

I want to prove the "only if" part with the same spirit, but there is something different. In the previous argument "$a\neq b \Rightarrow (\exists\alpha,<a,f^*\alpha>\neq<b,f^*\alpha>)$" is because there is a surjection $$H^k(X) \to Hom(H_k(X),R)$$ where $R$ is the coeffient ring. But now there may happen that $\{f^*\alpha\}$ does not contains all the values in $H^k(X)$, but enough to distinguish among $H_k(X)$

More precisely, there are two obstructions:

1. There may happen that $\{f^*\alpha\}$ takes all the values in the image $Hom(H_k(X),R)$, but not in $H^k(X)$.
2. There may happen that $\{f^*\alpha\}$ does not take all the values in the image $Hom(H_k(X),R)$, but enough to distinguish all those $H_k(X)$.

We need to avoid these phenomena. For 1 the UCT says the kernel of the map $H^k(X) \to Hom(H_k(X),R)$ is some $Ext^1(R\text{-module}, R)$, so I need to expect $R$ being injective. For 2, it is ok if it is of finite rank, or over some complete field (by geometric Hahn-Banach theorem). Hence everything is fine if $R$ is a complete field.

I want to know is this always true for arbitrary ring $R$?

Answer 1

This implication is not true in either direction. For instance, let $X=Y=\mathbb{RP}^2$ and let $f$ be a constant map. Then (with integer coefficients everywhere) we have $H_1(X)=H_1(Y)\cong\mathbb{Z}/2$ and $H^2(X)\cong H^2(Y)\cong\mathbb{Z}/2$, whereas $H^1$ and $H_2$ are trivial. So, $f^*:H^1(Y)\to H^1(X)$ is surjective but $f_*:H_1(X)\to H_1(Y)$ is not injective, while $f_*:H_2(X)\to H_2(Y)$ is injective but $f^*:H^2(Y)\to H^2(X)$ is not surjective.

Your argument for the "if" direction is incorrect because an element of $H_k(X)$ need not be determined by its images under all homomorphism $H_k(X)\to R$. That's exactly what happens in this example, with $R=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}/2$.