This is a question following p.16 of Scorpan's The Wild World of 4-Manifolds.

Consider a closed oriented 4-manifold $M$. Then $H_k(M; \mathbb{Z}) \cong F_k \oplus T_k$ for all $k$, where $F_k$ is free abelian and $T_k$ is torsion. By applying UCT and duality one can show that the homology and cohomology groups with $\mathbb{Z}$ coefficients are as follows: $$ \begin{align*} H_0(M) \cong H_4(M) \cong H^0(M) \cong H^4(M) &\cong \mathbb{Z} \\ H_1(M) \cong H^3(M) &\cong F_1 \oplus T_1 \\ H_2(M) \cong H^2(M) &\cong F_2 \oplus T_1 \\ H_3(M) \cong H^1(M) &\cong F_1 \end{align*} $$

My question concerns the (co)homology groups of $M$ with $\mathbb{Z}/p$ coefficients for $p$ prime. Since $\mathbb{Z}/p$ is a field, we need only calculate $H_i(M; \mathbb{Z}/p)$ for $i=0,1,2$ and then apply duality and UCT to determine all the desired groups. Since $M$ is connected, $H_0(M; \mathbb{Z}/p) \cong \mathbb{Z}/p$. To calculate $H_1(M; \mathbb{Z}/p)$, we consider the short exact sequence from UCT: $$ 0 \rightarrow H_1(M; \mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}/p \rightarrow H_1(M; \mathbb{Z}/p) \rightarrow \textrm{Tor}(H_0(M; \mathbb{Z}), \mathbb{Z}/p) \rightarrow 0. $$

Since $H_0(M; \mathbb{Z})$ is free abelian, the $\textrm{Tor}$ part vanishes, and we end up with $$ H_1(M; \mathbb{Z}/p) \cong F_1/pF_1 \oplus T_1/pT_1 =: F_1' \oplus T_1', $$ where $T_1'$ is the $p$-torsion (this claims follows from decomposing $T_1$ as a direct sum of cyclic groups and observing that $\mathbb{Z}/n \otimes \mathbb{Z}/m \cong \mathbb{Z}/\gcd(n,m)$).

To calculate $H_2(M; \mathbb{Z}/p)$, we again apply UCT to get a short exact sequence $$ 0 \rightarrow [H_2(M; \mathbb{Z}) = F_2\oplus T_1] \otimes_\mathbb{Z} \mathbb{Z}/p \rightarrow H_2(M; \mathbb{Z}/p) \rightarrow \textrm{Tor}([H_1(M; \mathbb{Z}) = F_1\oplus T_1], \mathbb{Z}/p) \rightarrow 0. $$ Now the first term, similarly to the above, is $F_2' \oplus T_1'$, while the $\textrm{Tor}$ term vanishes on the free direct summand and gives $\textrm{Tor}(T_1, \mathbb{Z}/p) \cong \ker(T_1 \overset{\times p}{\rightarrow} T_1)$, i.e., the $p$-torsion $T_1'$. Since the sequence (non-canonically) splits, we get that $$ H_2(M; \mathbb{Z}/p) \cong F_2' \oplus T_1' \oplus T_1'. $$

This contradicts the result in Scorpan (who states the result for $p=2$), which says that in fact $H_2(M; \mathbb{Z}/p) \cong F_2' \oplus T_1'$. I cannot spot a mistake in my algebra, so I would very much appreciate if anyone could confirm whether I am right or overlooking something obvious.


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