Preserving simple-connectedness under intersection complexes

Question

Given a simplicial complex $X$, and a family of its subcomplexes $\{U_i\}_{i\in I}$, we define the corresponding intersection complex to be the simplicial complex $X_U$ with vertex set $I$ where $A \subseteq I$ form a cell whenever $\bigcap_{i\in A} U_i \neq \emptyset$. If in addition $\{U_i\}_{i\in I}$ covers $X$, i.e. if $\bigcup_{i\in I} = X$, and each $U_i$ is connected, we call $X_U$ a cover complex of $X$.

The question is:

If $X$ is a (finite) simply connected simplicial complex, must each cover complex of $X$ also be simply connected?

Here is an incomplete proof sketch. The case where $X$ is homeomorphic to the unit disc in the plane is easier (proof?). Let $C= v_0 v_1 \ldots v_k=v_0$ be a cycle in the 1-skeleton of $X_U$. We can restrict our attention to the case where $C$ is induced, i.e. no edge of $X_U$ connects vertices at distance at least two along $C$, because those cycles generate $\pi_1(X_U)$. Notice that such a cycle gives rise to a cycle $C’$ in $X$, obtained by uniting paths between $v_{i-1} \cap v_{i}$ and $v_i \cap v_{1+1}$ for each $i$. Since $X$ is simply connected, there exists a continuous map $F: D^2 \to X$ such that $F$ restricted to $\mathbb{S}^1$ traverses $C’$. If $F$ is bijective, we can reduce to the case where $X$ is homeomorphic to the unit disc mentioned above. If not, I hope that the disc case can be generalised…

Example: The answer is positive when $X$ is a tree. In this case $X_U$ must be a flag complex, i.e. every $k$ clique of the 1-skeleton $G_U$ of $X_U$ spans a $k$-cell. Moreover, it is known that $G_U$ is a chordal graph, which implies that the triangles of $G_U$ generate $\pi(G_U)$. Since $X_U$ is flag, each triangle of $G_U$ spans a homeomorph of a disc, and it easily follows that $X_U$ is simply connected.

Remarks on notation: The term intersection complex is standard, but cover complex is made up. Has it been studied under a different name?

If we replace cover complex by intersection complex in the question the answer is easily no, as we can remove vertices to create holes.

Answer 1

Yes.

There is a map $X\to X_U$, taking each point $x$ to some point in the simplex corresponding to the set of all $i$ such that $x\in U_i$.

This map induces a surjection $\pi_1(X)\to \pi_1(X_U)$. To see this, take any loop in $X_U$. It can be taken to be an edge path running through vertices $i_0,i_1,\dots , i_k=i_0$. Lift this to a loop in $X$ by sending each $i_j$ to some point $p_j\in U_{i_j}$, sending the midpoint of $[i_j,i_{j+1}]$ to some point $q_j\in U_{i_j}\cap U_{i_{j+1}}$, sending the left half of the interval $[i_j,i_{j+1}]$ to a path in $U_{i_j}$ from $p_j$ to $q_j$, and sending the right half to a path in $U_{i_{j+1}}$ from $q_j$ to $p_{j+1}$.