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The universal coefficient theorem tells us that we have a s.e.s.

$$0 \to H_n(X) \otimes \mathbb{Z}_p \to H_n(X; \mathbb{Z}_p) \xrightarrow{\beta} \mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p) \to 0.$$

In particular, we can identify $\mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p) \cong \ker (H_{n-1}(X) \xrightarrow{\cdot p} H_{n-1}(X))$.

From the Bockstein homomorphism $\tilde{\beta}$ of the s.e.s.

$$0 \to \mathbb{Z} \xrightarrow{\cdot p} \mathbb{Z} \xrightarrow{\mathrm{mod}~p} \mathbb{Z}_p \to 0,$$

we obtain this snippet of the corresponding l.e.s. in homology:

$$\cdots \to H_{n}(X; \mathbb{Z}_p) \xrightarrow{\tilde{\beta}} H_{n-1}(X) \xrightarrow{\cdot p} H_{n-1}(X) \to \cdots,$$

so that we can infer $\mathrm{im}{\tilde{\beta}} = \ker (H_{n-1}(X) \xrightarrow{\cdot p} H_{n-1}(X)) \cong \mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p)$. Hence, $\beta$ and $\tilde{\beta}$ have the same image.

Now, my question is whether indeed $i \circ \beta = \tilde{\beta}$, where $i: \mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p) \hookrightarrow H_{n-1}(X)$ is the natural inclusion coming from the identification above.

I don't need a full explanation. A hint might be enough.

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    $\begingroup$ What happens to the Bockstein if $H_{n-1}(X)$ is isomorphic to $\mathbb{Z}/(p^k)$ for some $k > 1$? What happens to $i \circ \beta$ in this case? $\endgroup$ Dec 7, 2021 at 22:50
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    $\begingroup$ Then $\mathrm{Tor}(H_{n-1}(X),\mathbb{Z}_p) \cong \left<p^{k-1}\right> \cong \mathbb{Z}_p$, but how do I go on from here? $\endgroup$
    – YoungMath
    Dec 8, 2021 at 18:09
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    $\begingroup$ I think both $i$ and $\beta$ will be nonzero, and their composite will be nonzero. But the Bockstein operation should be zero in this case. $\endgroup$ Dec 8, 2021 at 19:49
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    $\begingroup$ For the Bockstein, we have this segment of the l.e.s.: $\cdots \to H_n(X;\mathbb{Z}_p) \xrightarrow{\tilde{\beta}} \mathbb{Z}_{p^k} \xrightarrow{\cdot p} \mathbb{Z}_{p^k} \to \cdots$ and hence $\mathrm{im}(\tilde{\beta}) = \ker(\mathbb{Z}_{p^k} \xrightarrow{\cdot p} \mathbb{Z}_{p^k}) = \left<p^{k-1}\right> \neq 0$. In fact, the image of $\tilde{\beta}$ is always $\mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p)$, and therefore $\tilde{\beta}$ and $\beta$ have the same image. Or do I overlook something obvious here? $\endgroup$
    – YoungMath
    Dec 8, 2021 at 21:08
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    $\begingroup$ Maybe you're answering your own question... $\endgroup$ Dec 9, 2021 at 5:59

1 Answer 1

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Here is what I've obtained so far.

$$0 \to C(X) \xrightarrow{\cdot p} C(X) \xrightarrow{\mod p} C(X) \otimes \mathbb{Z}_p \to 0$$

gives the long exact sequence in homology

$$\dots \to H_n(X) \xrightarrow{\cdot p} H_n(X) \xrightarrow{\mod p} H_n(X; \mathbb{Z}_p) \xrightarrow{\tilde{\beta}} H_{n-1}(X) \xrightarrow{\cdot p}H_{n-1}(X) \to \dots$$

From that, we obtain a s.e.s.

$$0 \to \mathrm{coker}(\_ \cdot p) \to H_n(X; \mathbb{Z}_p) \to \ker(\_ \cdot p) \to 0.$$

Where $\mathrm{coker}(\_ \cdot p) \cong H_n(X) \otimes \mathbb{Z}_p$ and $\ker(\_ \cdot p) \cong \mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p)$. Thus,

$$0 \to H_n(X) \otimes \mathbb{Z}_p \to H_n(X; \mathbb{Z}_p) \to \mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p) \to 0.$$

In that spirit, $\mathrm{Tor}(H_{n-1}(X), \mathbb{Z}_p)$ is obtained from the free resolution induced by $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_p \to 0$.

This should resemble the universal coefficient theorem.

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