I am considering the bockstein $\beta$ for $p=2$, so for the exact sequence $$0\rightarrow \mathbb Z_2\rightarrow \mathbb Z_4\rightarrow \mathbb Z_2\rightarrow 0$$ I would like to prove it is a stable cohomology operation, that is it commutes with the suspension isomorphism $$\sigma: H^n(X,\mathbb Z_2)\cong H^{n+1} (\Sigma X, \mathbb Z_2)$$ I tried to use the naturality of bockstein and $$H^n(X,\mathbb Z_2)\rightarrow H^n(\Sigma X,\mathbb Z_2)\xrightarrow{\beta}H^{n+1}(\Sigma X,\mathbb Z_2)$$ and similarly $$H^{n+1}(X,\mathbb Z_2)\rightarrow H^{n+1}(\Sigma X,\mathbb Z_2)\xrightarrow{\beta}H^{n+2}(\Sigma X,\mathbb Z_2)$$ I got that these 2 maps commute with bockstein. I think these maps may be the same as the suspension isomorphism but I really have no idea how that might be. I know the suspension isomorphism is obtained by Mayer-Vietoris, but the composition here uses the long exact sequence.
Does anyone have an idea of if that is true? If it is false, do you know how to prove that $\beta$ is stable? I found a general proof for Steenrod squares but I haven't seen that yet, so I would like to know is there's a simpler proof just for bockstein.
UPDATE: I proved that the commuting with the suspension isomorphism is equivalent to commuting with the coboundary $H^n(X, \mathbb Z_2)\rightarrow H^{n+1}(CX, X,\mathbb Z_2)$, but I don't see how to show that.