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I want to prove that $\dim_{\mathbb{Z}_2}(H_k(M;\mathbb{Z}_2))=\dim_{\mathbb{Z}_2}(H_{n-k}(M,\partial M;\mathbb{Z}_2))$.

This seems like a trivial result from Poincaré-Duality, so I tried \begin{align} H_{n-k}(M,\partial M;\mathbb{Z}_2)=&H^k(M;\mathbb{Z}_2)\\ =&Hom(H_k(M;\mathbb{Z}_2))+Ext(H_{k-1}(M;\mathbb{Z}_2);\mathbb{Z}_2) \end{align} by using PD and universal coefficient theorem. Now I am unsure of how to get rid of the $Ext$-functor. I know the result would fit if $H_{k-1}$ would be free, but since that is not given I am clueless and I am generally unsure how to approach the $\mathbb{Z}_2$-coefficients.

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Specifically the universal coefficient theorem says there is a short exact sequence

$$ 0\to Ext_R^1(H_{n-1}(X; R), G) \to H^n(X;G) \to Hom_R(H_n(X;R), G) \to 0 $$

for any principal ideal domain $R$ and $R$-module $G$. In your case both $R$ and $G$ are $\mathbb{Z}/2$, but $\mathbb{Z}/2$ is free as a module over itself so the $Ext$ term vanishes.

More to the point, whenever $\mathbb{F}$ is a field the term "module" means "vector space" so ALL modules are free, so the $Ext_{\mathbb{F}}$ term ALWAYS vanishes and $$H^n(X;G) \cong Hom_{\mathbb{F}}(H_n(X;\mathbb{F}), G) = Lin(H_n(X;\mathbb{F}), G)$$ for any $\mathbb{F}$-vector space $G$.

Edit: in light of some comments I should clarify that in the context of $R$-modules, "free" means "has a basis": https://en.wikipedia.org/wiki/Free_module. $\mathbb{Z}/2$ can be viewed as a $1$-dimensional vector space over itself.

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  • $\begingroup$ Thank you, a few questions for clarity with beeing free over a $G$: Is $\mathbb{Z}$ free over $\mathbb{Z}_2$? (My guess would no) Is being free over $\mathbb{Z}$ equivalent to beeing torsion free? (My guess would be yes) I'm sorry it's been a while since i heard algebra. :) $\endgroup$
    – Timmathy
    Commented May 31, 2020 at 21:13
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    $\begingroup$ To answer those questions, $\mathbb{Z}$ isn't really a $\mathbb{Z}/2$-module (or in this case, a $\mathbb{Z}/2$-vector space), so I'm not sure if asking that makes sense. Second, yes, being free over $\mathbb{Z}$ is exactly being torsion free. $\endgroup$
    – Osama Ghani
    Commented May 31, 2020 at 21:22
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    $\begingroup$ Here's how you can see $\mathbb{Z}$ admits no $\mathbb{Z}/2$-module structure: if $1_2$ denotes the multiplicative identity aka non-zero element of $\mathbb{Z}/2$, you would have to have $1_2 \cdot n = n$ for all $n$, but also $1_2\cdot n + 1_2\cdot n = (1_2 + 1_2)\cdot n = 0\cdot n = 0$. This argument shows that any $\mathbb{Z}/2$-module must be a group where every element is $2$-torsion, and in fact since $\mathbb{Z}/2$ is a field the term "module" is the same as "vector space" so every $\mathbb{Z}/2$ module is a direct sum of copies of $\mathbb{Z}/2$. $\endgroup$
    – William
    Commented May 31, 2020 at 21:29
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    $\begingroup$ You might be thinking of how $\mathbb{Z}$ has the structure of a module over the group ring $\mathbb{Z}[\mathbb{Z}/2]$, where the non-trivial element of $\mathbb{Z}/2$ acts via $-1$ in $\mathbb{Z}$. This module is certainly not free. (In the context of $R$-modules, "free" means "has a basis": en.wikipedia.org/wiki/Free_module) $\endgroup$
    – William
    Commented May 31, 2020 at 21:36
  • $\begingroup$ Thank you very much! Now things start to get clearer! $\endgroup$
    – Timmathy
    Commented May 31, 2020 at 21:38

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