The context of my question is to solve this problem, but my question is not really about the solution, it's about the definition and implications of the product topology.
Let $X,Y,Z$ be topological spaces and $A \in Op(X \times Y), B \in Op(Y\times Z)$
Show that $A \circ B = \{(x,z) \in X\times Z : \exists y \in Y s.t. (x,y) \in A, (y,z)\in B\}$ is an open set of $X \times Z$
NOTATION
$Op(\cdot)$ means the open sets of a space (i.e., the elements of the topology of such space).
$\tau_x,\tau_y,\tau_z$ are the topologies of the topological spaces $X,Y,Z$ respectively.
$\pi_x,\pi_y,\pi_z$ are the projections from the product spaces. e.g. $\pi_x:X\times Y \to X$. I'm using the same symbol $\pi_x$ for the functions from $X\times Y$ to $X$ and from $X \times Z$ to $X$.
$(\pi_x)^{-1}\tau_x$ is the inverse image of the topology $\tau_x$ by the projection $\pi_x$
$Sub(X) = \{A:A \subseteq X\}$
The product topology defined in $X \times Y$ is the topology generated by $((\pi_x)^{-1}\tau_x )\cup ((\pi_y)^{-1}\tau_y)$, i.e., the smallest topology that contains all the elements in $(\pi_x)^{-1}\tau_x$ and in $(\pi_y)^{-1}\tau_y$. (correct so far?)
Now, the set $(\pi_x)^{-1}\tau_x = \{A \in Sub(X \times Y): \forall (\alpha,\beta)\in A: \exists B \in \tau_x, \alpha \in B\}$ (is this also correct?)
If such definitions above are correct, can I make the following claim?
Let $\tau_{x \times y}$ be the product topology on $X \times Y$. Then $\forall A \in \tau_{x \times y}$ if $(\alpha,\beta) \in A \implies \exists B \in \tau_x s.t. \alpha \in B$ and/or $\exists C \in \tau_y s.t. \beta \in C$
Is this correct?(if it is, I believe the theorem I'm supposed to prove is false because I could find sets $A$ and $B$ where only the coordinate with "$y$" would be in a set in $\tau_y$ and the other coordinates wouldn't be in a set of the topologies of $X$ and $Z$ and so the set $A \circ B$ wouldn't be in the product topology of $X \times Z$)