How to show $\mathbb{CP}^{2}\#\overline{\mathbb{CP}^{2}}\not\cong \mathbb{S}^{2}\times \mathbb{S}^{2}$? I was asked to prove that $$\mathbb{CP}^{2}\#\overline{\mathbb{CP}^{2}}\not\cong \mathbb{S}^{2}\times \mathbb{S}^{2}$$ as fibre bundles over $\mathbb{S}^{2}$ with fibre $\mathbb{S}^{2}$. Since the above connected sum is as manifolds instead of as bundles, it is not clear to me how to prove it via elementary methods. 
Here are some thoughts:
1) Can I prove them non-homotopic as manifolds? If they are homotopic as fibre bundles, then they should be homotopic as manifolds as well. Therefore, since all the algebraic invariants on the right hand side is relatively easy to compute, I should be able to solve the problem. However, it is not clear to me how the underlying holomorphic structure of $\overline{\mathbb{CP}^{2}}$ influence the structure. For example, since $\mathbb{CP}^{2}, \mathbb{S}^{2},\overline{\mathbb{CP}^{2}}$ are all simply connected, we cannot get any information from $\pi_{1}$. For $\pi_{2}$ we know $\pi_{2}(\mathbb{S}^{2}\times \mathbb{S}^{2})=\pi_{2}(\mathbb{S}^{2})\times \pi_{2}(\mathbb{S}^{2})=\mathbb{Z}\times \mathbb{Z}$ via Hurewitz's theorem. But computing $$\pi_{2}(\mathbb{CP}^{2}\#\overline{\mathbb{CP}^{2}})$$ seem to be nontrivial even if one use Ryan Budney's method (https://mathoverflow.net/questions/93282/homotopy-groups-of-connected-sums) because they are both simply connected. We can also compute the homology, but it is not clear to me what kind of space is $\mathbb{CP}^{2}-\mathbb{D}^{4}$. If we view $\mathbb{CP}^{2}\cong \mathbb{CP}^{1}\cup \mathbb{C}^{2}$ with the attaching map given by $\mathbb{S}^{3}\rightarrow \mathbb{S}^{2}$, then remove $\mathbb{D}^{4}$ seems to be giving us $\mathbb{CP}^{1}$ back. But this is not rigorous. 
Using the hints at here (Computing the homology and cohomology of connected sum) it seems at $H_{2}$ level they again coincide: the result is $\mathbb{Z}\oplus \mathbb{Z}$. 
The most important thing is I noticed I did not use the conjugate relationship in the above arguments at all. So there must be something missing here. I guess I can try to compute $\pi_{3}$ and $\pi_{4}$ as well. Since they are CW Complexes, they must differ at some point otherwise it would violate Whitehead's thoerem. 
2) Can I prove this via cohomological methods like Chern class? Again, I do not know how to compute it...
 A: It seems that the intersection form on second cohomology is indeed enough to distinguish these 4-manifolds, as long as you use integer coefficients. (I'm not sure why I did not think of that earlier, since by Freedman's results this is the first thing one should check.) 
All cohomology is with $\mathbb{Z}$ coefficients. Your manifolds are orientable, so a choice of orientation on each of them induces an intersection form 

$H^{2}(M) \times H^{2}(M) \rightarrow H^{4}(M) \rightarrow \mathbb{Z}$

given by cup product and then evaluation on the fundamental class. (See: Wikipedia - Intersection form).
The intersection form of $\mathbb{CP}^{2}$#$(-\mathbb{CP}^{2})$ is, up to a sign, given by 
$\left( \begin{array}{ccc}
1 & 0 \\
0 & -1 \end{array} \right)$,
intuitively since the square of the generator of $H^{2}(\mathbb{CP}^{2})$ is equal to $1$ and square of the generator of $H^{2}(-\mathbb{CP}^{2})$ is -1 because of the reversed orientation.
On the other hand, the intersection form of $S^{2} \times S^{2}$ is, up to a sign, given by
$\left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \end{array} \right)$,
since if $h_{1}, h_{2}$ are the generators of $H^{2}(S^{2})$ of the first, resp. second copy of $S^{2}$, then $h_{1} \cup h_{2}$ is the generating $4$-cell of $H^{4}(S^{2} \times S^{2})$, but $h_{i}^{2} = 0$ for dimensional reasons. Since a general element of $H^{2}(S^{2} \times S^{2})$ can be written as $ah_{1} + bh_{2}$ with $a, b \in \mathbb{Z}$ and 

$(ah_{1} + bh_{2})^{2} = a^{2}h_{1} ^{2} + b^{2} h_{2}^{2} + 2ab h_{1} h_{2} = 2ab$

one sees that there is no element of $H^{2}(S^{2} \times S^{2})$ that squares to an odd number, in stark contrast with the intersection form on $\mathbb{CP}^{2}$#$(-\mathbb{CP}^{2})$. This proves that these two manifolds are not even homotopy equivalent, so in particular not diffeomorphic.
