Use intersection theory to show that a closed 4-manifold is not homotopy equivalent to $S^2\times S^2$. Let $R_\theta: S^2\to S^2$ be the counterclockwise rotation around $z-$axis. Define $M$ to be the closed 4-manifold obtained by gluing two copies $A_1,A_2$ of $B^2\times S^2$ along their common boundary $S^1\times S^2$. The gluing map is $(e^{i\theta},v)\to(e^{i\theta},R_{\theta}(v))$. The problem asks to 
i) compute the homology of $M$;
ii) find another closed 4-manifold $N$ which has the same homology and show that it is not homotopy equivalent to $M$ by using intersection theory.
For part i), I have found that $M$ has the same homology as $S^2\times S^2$. I'm stuck with part ii).
My Questions:

*

*I have noticed that $M$ is the nontrivial bundle $S^2\tilde{\times}S^2$ so it remains to compare the intersection forms of $S^2\times S^2$ and $S^2\tilde{\times}S^2$. Using cup product I managed to show that the former can be represented by a matrix \begin{bmatrix}0&1\\1&0\end{bmatrix} but how to compute that of the nontrivial one?

*Since I didn't know how to compute the intersection form, I tried to find some other approach to avoid that. Suppose $I_2$ is the intersection form mod 2 of $S^2\times S^2$. Because the matrix of $I_2$ is known, $I_2(\alpha,\alpha)=0$ for any homology class $\alpha$. In particular, the self intersection number mod 2 of any 2 dimensional submanifold is 0. Therefore, it would be done if there exists some 2 dimensional submanifold of $M$ whose self intersection number mod 2 is 1. Is there any simple way to construct such a submanifold?

 A: I'm going to view $S^2\subseteq \mathbb{C}\times \mathbb{R}$, imagining that the rotation occurs in the $\mathbb{C}$ portion (so $\{0\}\times\mathbb{R}$ is the $z$-axis.)
For $i=1$ and $i=2$, let $N_i\subseteq A_i$ be defined as $N_i = B\times \{(0,1)\}$.
Suppose $p = (e^{i\theta}, (0,1))\in \partial N_i$, the boundary of $N_i$.  Then the gluing map (which I'll call $f$) is $f(e^{i\theta},(0,1)) = (e^{i\theta}, (0,1))$, so we can view $f$ as a homeomorphism $\partial N_1\rightarrow \partial N_2$.
In particular, we can glue $N_1$ to $N_2$ to obtain a closed manifold $N$.
Proposition  The manifold $N$ has self intersection number $\pm 1$.
To prove this, we'll first construct a deformation of $N$.  To that end, for fixed small real number $t\in (-\varepsilon,\varepsilon)$, consider the subset $N_1(t) = B\times \{(\sin(t),\cos(t)\}$.
Note that $N_1(0) = N_1$, so this is a deformation of $N_1$.  At the boundary $\partial N_1(t)$, the gluing map $f$ takes the form $f(e^{i\theta}, (\sin(t),\cos(t)) = (e^{i\theta}, (e^{i\theta} \sin(t),\cos(t)))$.
Now, consider $N_2(t) = \{(re^{i\theta}, (re^{i \theta}\sin(t), \sqrt{1-r^2\sin^2(t))} \}$.
Note that $N_2(0) = N_2$, and that $\partial N_2(t)$ (which occurs when $r=1$) is given by $\partial N_2(t) = \{(e^{i\theta}, e^{i\theta} \sin(t), \cos(t))$}.  That is, $f:\partial N_1(t)\rightarrow \partial N_2(t)$, so for each $t$ we can glue to obtain a smoothly embedded submanifold $N(t)$.
Having defined $N$ and $N(t)$, we can now compute the intersection $N\cap N(t)$.
Proposition:  For $t\neq 0$, $N\cap N(t) = \{ 0,(0,1)\}$.
Proof:  In $N$, every point has last coordinate $(0,1)$.  So we need to determine when this happens for $N(t)$ with $t\neq 0$.
In $N_1(t)$, the last coordinate is $(\sin(t),\cos(t))$, so $N_1(t)$ is disjoint from $N$ for all $t\neq 0$.  In $N_2(t)$, since $t\neq 0$, we find that the last coordinate is $(0,1)$ iff $r = 0$.
When $r=0$, we obtain the point $(0,0,1)\in N_2\cap N_2(t)$.  $\square$
Now that we've computed the intersection, we need only show that $N$ and $N(t)$ intersect transversely at $(0,0,1)$.
Proposition  The intersection at $(0,0,1)$ of $N$ and $N(t)$ is transverse.
Proof: Observe that $T_{(0,(0,1))} B\times S^2$ can be identified by $T_0 B\oplus T_{(0,1)}S^2\cong \mathbb{R}^2\oplus \mathbb{C}$.
Then $T_{(0,(0,1))} N = \mathbb{R}^2\oplus 0$, obviously.
Now, consider the curve $\alpha(s) = (s, s\sin(t), \sqrt{1-s^2 \sin^2(t)})$.  This curve lies in $N(t)$ (take $t=s, \theta = 0$).  Then $\alpha'(0)$, when projected to $\mathbb{C}$, is $\sin(t)$, which is a non-zero real number for $t\neq 0$.
Lastly, consider the curve $\beta(s) = (ise, is\sin(t), \sqrt{1-s^2 \sin^2(t)})$.  This curve lies in $N(t)$ (take $t=s, \theta = \pi/4$).  Then $\beta'(0)$, when projected to $\mathbb{C}$, is $i\sin(t)$, which is a non-zero purely imaginary number for $t\neq 0$.
Thus, we see the intersection is transverse. $\square$
Now that we have a unique point of intersection, and the intersection is transverse, it follows that $N$ has self intersection number $\pm 1$, as claimed.
