The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely, $$ \mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x. $$ The $\cup_1$ is a higher cup product. The $Sq^1 x= x \cup_1 x$. It shall be true that $$ \mathcal{P}(x) \mod 2= x \cup x. $$

Question 1: If $ x \cup x =0 \mod 2$, and if $ x \cup_1 Sq^1 x =0 \mod 2$, is it true that $$ \mathcal{P}(x) =0 \mod 4? $$

- If not, please provide some counter examples.

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Question 2: $\mathcal{P}(x)$ is a well-defined invariant for the cobordism $\Omega^4_{SO}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4$. Is $$ \frac{1}{2}(\mathcal{P}(x) -x^2) \mod 2 = x \cup_1 Sq^1 x \mod 2 $$ a well-defined invariant of the cobordism $\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2$?

Question 3: Please provide the correct way to write the cobordism generator of $$\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2.$$