A while ago, I thought of the following question in relation to some other work I was doing mathematically:
Let $f_1,f_2,f_3:\mathbb C^2\rightarrow \mathbb C$ be three linear maps and $A_1,A_2,A_3\subseteq\mathbb C$ be three simply connected open subsets of $\mathbb C$. Let $B_i=f_i^{-1}(A_i)$. Is it necessarily the case that $B_1\cap B_2\cap B_3$ is a union of simply connected components?
I've had very little luck in considering this question or even with weaker versions of it; if we only had two sets or if we were in only one dimension, it's not too hard to prove a positive answer to this question. Similarly, if the sets were convex, the answer is clearly positive.
My initial thought was to try to get information about this intersection by looking at the Meyer-Vietoris sequence of pairs such as $(B_1,B_2\cap B_3)$ or $(B_1\cap B_2, B_2\cap B_3)$, but these sequences involve the homology of sets such as $B_1\cup (B_2\cap B_3)$ which I can only think to study by looking at the sequence for the pair $(B_1\cup B_2, B_1\cup B_3)$, which then involves the set $B_1\cup B_2\cup B_3$ which I also don't have a good way to understand - basically, I can get a whole web of morphisms between relevant sets, but never any results. I've also thought to try using something like Lefschetz duality, but this never seems to lead anywhere.
Another interesting approach is to fix a loop $\gamma$ in $\mathbb C^2$ and then look at the intersection of every such cylinder containing $\gamma$ to see if $\gamma$ could be contracted within any open set containing that intersection - but these intersections seem really difficult to reason about.
I don't have a strong sense of whether the answer is positive or negative - my work on the question leaves me pretty unsure of it. Is there a nice proof or counterexample to the question? I'd also be curious if, in the event that the answer is positive, whether the answer is the same for more cylinders or more dimensions.
No. Choose \begin{align*} f_1(z,w)&=z\\ f_2(z,w)&=w\\ f_3(z,w)&=z+w\\ U_1&=\{z\in\mathbb C\mid \operatorname{Im} z<(\operatorname{Re}z)^2\}\\ U_2&=U_1\\ U_3&=\{z\in\mathbb C\mid \operatorname{Im} z>1\} \end{align*}
Then \begin{align*} B_1&=\{(z,w)\in\mathbb C^2\mid \operatorname{Im} z<(\operatorname{Re}z)^2\}\\ B_2&=\{(z,w)\in\mathbb C^2\mid \operatorname{Im} w<(\operatorname{Re}w)^2\}\\ B_3&=\{(z,w)\in\mathbb C^2\mid \operatorname{Im}(z+w)>1\}. \end{align*}
If $(z,w)\in B_1\cap B_2\cap B_3$ then $(\operatorname{Re}z)^2+(\operatorname{Re}w)^2>1,$ so in particular $(\operatorname{Re}z,\operatorname{Re}w)\neq(0,0).$ Define $h:\mathbb C^2\to\mathbb R^2\setminus\{(0,0)\}$ by $$h(z,w)=(\operatorname{Re}z,\operatorname{Re}w).$$
Define $\gamma:\mathbb R/2\pi \mathbb Z\to B_1\cap B_2\cap B_3$ by $$\gamma(\theta)=(2\sin\theta+2i\sin^2\theta,2\cos\theta+2i\cos^2\theta).$$ Then $h(\gamma(t))=(2\sin\theta,2\cos\theta)$ winds once around $(0,0)$ as $t$ goes from $0$ to $2\pi.$ This shows that $$h^*:\pi_1(B_1\cap B_2\cap B_3,\gamma(0))\to\pi_1(\mathbb R^2\setminus\{0,0\},h(\gamma(0)))$$ has a non-trivial image, which implies that its domain is non-trivial.