Inverse of projection map in product topology which defines sub-basis.

Question

I have $(X,\tau_X)$ and $(Y,\tau_Y)$ – two topological spaces. Consider $U\subseteq X$ and $V\subseteq Y$ and two projection maps $\pi_X: X\times Y \to X$ and $\pi_Y: X \times Y \to Y$. Then it is well known that $$\mathcal B = \{\pi_X^{-1}(U): U \in \tau_X\} \cup \{\pi_Y^{-1}(V): V \in \tau_Y\}$$ is a sub-basis for the product topology on $X\times Y$. I am wondering about that inverse mapping because I can take $U_1 \times V_1$ and $U_1 \times V_2$ then use $ \pi_X(X \times V) $ and I have $ U_1 $ which is open. Then I am using $ \pi_X^{-1}(U) $ an I have ambiguity in result as I can have two results $ U_1 \times V_1 $ and $ U_1 \times V_2 $, then inverse is not map. Where my logic fails, point me please?

Answer 1

By definition of the inverse mapping for $U \in \tau_X$,

$$\pi_X^{-1}(U) = \{(x,y) \in X \times Y \mid \pi_X(x,y) \in U\} = U \times Y.$$

So I don't really see where the ambiguity is. To be more specific, you don't have the equality $\pi_X^{-1}(U) = U \times V$ for $V \subsetneq Y$.