Suppose $M$ en $N$ are differentiable manifolds with differentiable structures $\{(U_a,x_a)\}$ and $\{(V_b,x_b)\}$ resp. Consider $M\times N$ and the mappings $z_{ab}(p,q):=(x_a(p),y_b(q))$ with $p\in U_a$ and $q\in V_b$.
I want to prove that $\{(U_a\times V_b,z_{ab})\}$ gives a differentiable structure on $M\times N$ such that the projections $\pi_1$ and $\pi_2$ are differentiable.
Is this good?!: $$\bigcup_{ab}{z_{ab}(U_a\times V_b)}=\bigcup_a{x_a(U_a)}\times\bigcup_b{y_b(V_b)}=M\times N$$ My problem lies in the second point of the definition of a differentiable manifold: Suppose $z_{ab}(U_a\times V_b)\cap z_{cd}(U_c\times V_d)=W\neq\emptyset$ Why are $z^{-1}_{ab}(W)$ and $z^{-1}_{cd}(W)$ open?
I think $z^{-1}_{cd}\circ z_{ab}$ is differentiable because: $$z^{-1}_{cd}\circ z_{ab}(p,q)=z^{-1}_{cd}(z_{ab}(p,q))=((x^{-1}_c\circ x_a)(p),(y^{-1}_d\circ y_b)(q)$$ and this is differentiable per assumption in each component thus differentiable.
But why are the projections differentiable? Herefore we have to prove that $y^{-1}_b\circ\pi\circ x_a$ are differentiable for all $a$ and $b$ right?! But how to do this?
Thank you for help :)
Your definition of the differentiable structure is fine. (EDIT: Let me expand that a bit, I haven't recognised your first question the first time.) Indeed, since the families of open sets $\{x_a(U_a)\}_a$ cover $M$ and $\{y_b(V_b)\}_b$ covers $N$, their pairwise products form an open covering of $M\times N$. Even though it's true in that case (because we take all possible combinations), I wouldn't simply write $$\bigcup_{a,b}z_{a,b}(U_a\times V_b) = \bigcup_a x_a(U_a)\times\bigcup_b y_b(V_b).$$ It's $\bigcup_{a,b}z_{a,b}(U_a\times V_b) = \bigcup_{a,b}x_a(U_a)\times y_b(V_b)$, thus the only trivial inclusion is $$\bigcup_{a,b}z_{a,b}(U_a\times V_b)\subset \bigcup_a x_a(U_a)\times\bigcup_b y_b(V_b)=M\times N.$$ To see equality, given any point $(p,q)\in M\times N$, there exist indices $a,b$, such that $p\in x_a(U_a)$ and $q\in y_b(V_b)$. Hence, $(p,q)\in x_a(U_a)\times y_b(V_b)\subset\bigcup_{a,b}z_{a,b}(U_a\times V_b)$.
Now, given $U_a$, $V_b$, $U_c$ and $V_d$, $$W=z_{a,b}(U_a\times V_b)\cap z_{c,d}(U_c\times V_d) = (x_a(U_a)\times y_b(V_b))\cap(x_c(U_c)\times y_d(V_d))$$ is an intersection of (products of) open sets, hence an open subset of $M\times N$. The maps $x_a$ and $y_b$ are continuous, and therefore so is their product $z_{a,b}$. Consequently, $z_{a,b}^{-1}(W)$ is an open subset of $x_a(U_a)\times y_b(V_b)$.
The expression $y_b^{-1}\circ\pi\circ x_a$ isn't well defined. ($x_a$ maps to $M$, so there is no $\pi$ to apply.) To verify differentiability of the projections, the maps $x_{a'}^{-1}\circ\pi_1\circ z_{a,b}$ and $x_{b'}^{-1}\circ\pi_2\circ z_{a,b}$ (where ever sensible) have to be proven to be differentiable.