Let $M$ be a path-connected finite $CW$-complex. Suppose the first integral homology group is $H_1(M;\mathbb{Z})= \mathbb{Z}_2^{\oplus r}\oplus A$ where $r\geq 1$ and $A$ is a finite abelian group of odd order. Does $M$ necessarily satisfy the property


For any nonzero element $x\in H^1(M;\mathbb{Z}_2)$, $x^2\neq 0$?

My attempt: I notice that for any nonzero element $x\in H^1(M;\mathbb{Z}_2)$, $x^2=Sq^1 x=\beta x$ where $\beta$ is the Bockstein homomorphism associated with the coefficient sequence $$ 0\to \mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}\to 0.$$ I want to use these formulas and prove $$ \text{Ker} \beta=0. $$ By the universal coefficient theorem, $$ H^1(M;\mathbb{Z}_4)=Hom(H_1(M;\mathbb{Z});\mathbb{Z}_4) =Hom(\mathbb{Z}_2^{\oplus r};\mathbb{Z}_4) =\mathbb{Z}_2^{\oplus r},$$

$$ H^1(M;\mathbb{Z}_2)=Hom(H_1(M;\mathbb{Z});\mathbb{Z}_2) =Hom(\mathbb{Z}_2^{\oplus r};\mathbb{Z}_2) =\mathbb{Z}_2^{\oplus r}.$$ By the construction of Bockstein homomorphism, we have an exact sequence $$ H^1(M;\mathbb{Z}_4)\overset{f}{\longrightarrow} H^1(M;\mathbb{Z}_2)\overset{\beta}{\longrightarrow }H^2(M;\mathbb{Z}_2).$$

Question: Why $\text{Ker}\beta=0$, which is equivalent to say that the image of $f$ is zero?

  • 3
    $\begingroup$ Did you try universal coefficients? Everything's in $H_1(X)$: $x^2=0$ iff $x$ lifts to $H^1(X;\mathbb{Z}/4)$. Hence, in your example, everything is in the parity of $n$. $\endgroup$ – Alex Degtyarev Mar 10 '16 at 12:32
  • 1
    $\begingroup$ Try using $Sq^1 = \rho\circ\beta$ where $\beta$ is the Bockstein associated to $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$ and $\rho$ is reduction mod 2. $\endgroup$ – Mark Grant Mar 10 '16 at 14:12
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    $\begingroup$ If $M$ has only finitely many 1-cells, (1) holds iff $H_1(M)\cong (\mathbb{Z}/2)^ r\oplus A$ for some $r \ge 0$ and $A$ a finite abelian group of odd order. $\endgroup$ – tj_ Mar 10 '16 at 18:13
  • $\begingroup$ @tj_ Thanks, Professor! I do not know how to prove your statement. Could you explain more? $\endgroup$ – QSH Mar 11 '16 at 5:42
  • 1
    $\begingroup$ Hint: the universal coefficient theorem is natural with respect to the coefficients. $\endgroup$ – Alex Degtyarev Mar 11 '16 at 8:46

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