Let $X$ be a simplicial complex and let $A \subset X$ be a contractible subcomplex on the same set of vertices as $X$. Is it true that the union $$\bigcup C$$ taken over all complexes $A \subset C \subset X$ whose inclusion in $X$ induces the trivial map on fundamental groups, has trivial fundamental group?



Does the union of the subcomplexes of $X$ that contain a given subcomplex and whose inclusion in $X$ is trivial on $\pi_1$, have trivial $\pi_1$?
I think it does not. Begin with $A$ a path $\{1,2\},\{2,3\},\{3,4\},\{4,5\}$. To construct $X$, add an edge $\{1,3\}$ and triangles $\{1,2,5\},\{1,3,5\},\{2,3,5\}$ to $A$.


$A$ is contractible. If $C$ contains $A$ and its image in $X$ has trivial fundamental group, it does not contain any of the edges $\{1,5\},\{2,5\},\{3,5\}$: for instance, if it contains $\{3,5\}$, then it contains the closed loop $3\to 4 \to 5 \to 3$ (a triangle) but there is no triangle containing the vertex $4$, so there is no nullhomotopy.


The path $1\to 2\to 3\to 1$ is contractible in $X$: the union of triangles we added is a disk which bounds it. Thus $\bigcup C$ of the question is just $A \cup \{1,3\}$. It contains none of the triangles of the complex: they each contain one of the forbidden edges $\{i,5\}$ where $i\in \{1,2,3\}$. So the path $1\to 2\to 3$ is not contractible in $\bigcup C$.


