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.
> 
> 
> 


--



> 
> **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 $$\Omega^4\_{Spin}(B^2 \mathbb{Z}\_2)=\mathbb{Z}\_2.$$
> 
> 
> 



Pontryagin square and $\frac{1}{2}(\mathcal{P}(x) -x^2) =x \cup_1 Sq^1 x$
(1) No, not for any reasonable interpretation of your condition "$x \cup\_1 Sq^1 x = 0$". Consider $M= S^2 \times S^2$, let $y \in H^2(M; \mathbb{Z})$ be the sum of the two obvious generators, and let $x \in H^2(M; \mathbb{Z}\_2)$ be the mod 2 reduction of $y$. Then $P(x)$ is the mod 4 reduction of $y^2 = 2$, but certainly $x^2 = 0$ mod 2 and $Sq^1 x = 0$.


(2) No, I don't think either side of your equation is a well-defined element of $H^4(M; \mathbb{Z}\_2)$ in general. Rather, Wu's theorem implies that if $M$ a closed spin 4-manifold and $x \in H^2(M;\mathbb{Z}\_2)$ then
$x^2 = Sq^2x = w\_2 x = 0$ (mod 2), so since the mod 2 reduction of $P(x)$ equals $x^2$ we get that $P(x) \in 2\mathbb{Z}\_4$.
Hence $\frac12 P(x)$ can be interpreted as a well-defined homomorphism $\Omega\_4^{Spin}(B^2\mathbb{Z}\_2) \to \mathbb{Z}\_2$.


(3) Taking $M = S^2 \times S^2$ and $x$ as in (1) shows that $\frac12 P(x)$ is non-trivial.


