TITLE: How to show $CP^1\times CP^1$ and $CP^2$ blow up one point are not diffeomorphic?
QUESTION [5 upvotes]: How can we prove $CP^1\times CP^1$ and $CP^2$ blow up one point is not diffeomorphic?
I tried to compute their Hodge numbers and Chern numbers but they are the same.

REPLY [3 votes]: You can do this using Kirby calculus and diagrams. I'll not attempt to draw any diagrams, but since you already have two clearly articulated answers, I'll just leave this as an alternative method.
Both the spaces are $S^2$ bundles over $S^2$, and as such are characterised up to diffeomorphism by their mod 2 Euler numbers. The Kirby calculus for such objects is well understood and documented.
Each space has a handle decomposition with exactly two 2-handles.  A Kirby diagram for $\mathbb{C}P^1\times\mathbb{C}P^1\cong S^2\times S^2$ consists of Hopf link of two $0$-framed unknots.
On the other hand a Kirby diagram of the blow up $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$ consists of a pair of disjoint unkots, one with framing $+1$ and one with framing $-1$. Performing a handle slide of the $-1$-unknot over the other generates a Hopf link of a $1$-framed unknot and a $-1+1=0$-framed unkot. 
Further handle slides can change the $(0,0)$-Hopf link of $S^2\times S^2$ into a $(0,2n)$-framed Hopf link. Similarly, handle slides can change the $(0,1)$-framed Hopf link of $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$ into a $(0,2n+1)$-framed Hopf link, but the difference of parity cannot be reconciled by any permitted Kirby moves.
Since the handle slide operation preserves the diffeomorphism type, it's clear now that the two spaces we started with are not diffeomorphic.