Inverse mapping on a set $U_1\times U_2$, wrong intuition? 
Let $f(x) = (f_1(x),f_2(x))$ where $f: X\to Y_1\times Y_2$. And $f_1:X\to Y_1, f_2: X\to Y_2$ where $X,Y_1,Y_2$ are topological spaces.

I want to prove some continuity properties, but my intuition gets stuck at a specific point.
Let $U_1$ be a neighbourhood of $f_1(x)\in Y_1$. Then $U_1\times Y_2$ will be a neighbourhood of $f(x) = (f_1(x),f_2(x)) \in Y_1\times Y_2$.
Consider $f^{-1}(U_1\times Y_2)$.
According to my notes $$f^{-1}(U_1\times Y_2) = f_1^{-1}(U_1) {\;\color{red}\cap\;} f_2^{-1}(Y_2)$$
I feel like there should be $\cup$. Could someone provide a simple intuitively argument on why there is a $\cap$. (Perhaps with a picture, I fail trying to visualize this)
PS: I have proven $f^{-1}(U_1\times Y_2) = f_1^{-1}(U_1) \cap f_2^{-1}(Y_2)$ using double inclusion, but I still feel like I'm not really getting it.
 A: More generally, for any $A\subseteq Y_1$ and $B\subseteq Y_2$, we have
$$f^{-1}[A\times B]=f_1^{-1}[A]\cap f_2^{-1}[B]\;.$$
The reason is that in order for $f(x)$ to be in $A\times B$, two things must both happen: we must have $f_1(x)\in A$, and we must have $f_2(x)\in B$. Having $f_1(x)\in A$ of course simply means having $x\in f_1^{-1}[A]$. Similarly, having $f_2(x)\in B$ means having $x\in f_2^{-1}[B]$. We need both of these things to happen, so we need $x$ to be in both of the sets $f_1^{-1}[A]$ and $f_2^{-1}[B]$. In other words, we need to have $x$ in the intersection of these two sets.
Replacing intersection by union would only ensure that $x$ was in at least one of the sets $f_1^{-1}[A]$ and $f_2^{-1}[B]$: if $x\in f_1^{-1}[A]\cup f_2^{-1}[B]$, then $f_1(x)\in A$ or $f_2(x)\in B$ (or both, of course). Having $f_1(x)\in A$ says nothing about $f_2(x)$: it implies only that $f(x)\in A\times Y_2$. Similarly, $f_2(x)\in B$ says only that $f(x)\in Y_1\times B$, since it places no restriction on $f_1(x)$. 
Thus, knowing that $x\in f_1^{-1}[A]\cup f_2^{-1}[B]$ tells you that 
$$f(x)\in(A\times Y_2)\cup(Y_1\times B)\;,$$
but it cannot guarantee that $x\in A\times B$. Indeed,
$$f^{-1}[(A\times Y_2)\cup(Y_1\times B)]=f_1^{-1}[A]\cup f_2^{-1}[B]\;.$$
The figure below shows three copies of $Y_1\times Y_2$; $Y_1$ is along the horizontal axis, and $Y_2$ along the vertical. In the first I’ve marked the set $A\subseteq Y_1$ and the product set $A\times Y_2$; you can see that $f(x)$ is in the vertical strip iff $f_2(x)\in A$. Similarly, the second copy shows $B\subseteq Y_2$ and the horizontal strip $Y_1\times B$; it should be fairly easy to see that $f(x)$ is in the strip iff $f_1(x)\in B$. Finally, the third copy shows $A\times B$: in order for $f(x)$ to be in that box, $f_1(x)$ must be in $A$, and $f_2(x)$ must be in $B$. In other words, $x$ must be in the intersection of $f_1^{-1}[A]$ and $f_2^{-1}[B]$. Allowing $x$ to be in their union would allow $f(x)$ to be anywhere in either of the strips in the first two copies – i.e., in a cross-shaped region in this schematic.

A: For $f$ to map a given $x \in X$ to $U_1 \times Y_2$, $x$ has to be in both the $f_1$ pre-image of $U_1$ and the $f_2$ pre-image of $Y_2$.  Logical AND (that is, conjunction), therefore intersection.
Example: Let $f: \mathbb{R} \to \mathbb{R}^2$ be defined by $f(\theta) = (\cos\theta, \sin\theta)$.  Note that $X, Y_1, Y_2$ are all represented by $\mathbb{R}$.  Let $\theta = 0$, and then $U_1$ is a neighborhood of $\cos 0 = 1$.  Clearly, $U_1 \times Y_2$ (which is a narrow strip of the $x$-$y$ plane between $x = 1-\varepsilon$ and $x = 1+\varepsilon$) has as its $f$ pre-image the cosine pre-image of $U_1$ intersected with the sine pre-image of $Y_2 = \mathbb{R}$—in other words, the intersection of $(1-\varepsilon, 1+\varepsilon)$ with $\mathbb{R}$, or $(1-\varepsilon, 1+\varepsilon)$ itself.
It certainly can't be the union, which would be all of $\mathbb{R}$.
