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?

1$\begingroup$ Note that by the van Kampen theorem, the set of "$C$"'s is closed under binary unions. It follows that the set of "$C$"'s is directed under inclusion, and therefore since $\pi_1$ commutes with directed unions, we have that $\pi_1(\bigcup C) \to \pi_1(X)$ is 0, i.e. $\bigcup C$ is itself a "$C$". In other words, the set of "$C$"'s has a unique maximal element, given by $\bigcup C$. $\endgroup$– Tim Campion ♦Oct 15, 2020 at 22:06

1$\begingroup$ And moreover, the "$C$"'s are closed under subcomplexes containing $A$, so a subcomplex of $X$ is a "$C$" if and only if it is a subcomplex of $\bigcup C$ containing $A$. $\endgroup$– Tim Campion ♦Oct 15, 2020 at 22:27

2$\begingroup$ (Let me just note a subtle point of the above argument: in order to use the van Kampen theorem, the pieces we are unioning must have a common basepoint. This is ensured by the condition that $A$ contain all the vertices.) $\endgroup$– Tim Campion ♦Oct 16, 2020 at 20:34

1$\begingroup$ What you have written is correct. I should also note that there is a connection with covering complexes: The union $\bigcup C$ is equal to the full subcomplex of the universal covering complex $\tilde{X}$ of $X$ on the vertices in the image of a lift of $A$ to $\tilde{X}$. $\endgroup$– erybOct 17, 2020 at 8:04
2 Answers
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 $\bigcup C$.

$\begingroup$ If I'm drawing it right, it looks to me like $X$ is basically a sphere with two discs removed, so homotopy equivalent to $S^1$, right? In addition to the edges you mention, the only other edges are $\{2,5\}$ and $\{3,5\}$ and $\{1,5\}$, right? It looks to me like $1 \to 2 \to 3 \to 1$ is not contractible  it looks like we would need a triangle $\{3,4,5\}$ to make it contractible. So it looks to me like $\bigcup C = A$ is contractible. $\endgroup$– Tim Campion ♦Oct 15, 2020 at 22:03

$\begingroup$ @TimCampion The union of the triangles in $X$ is a disk with boundary path $1\to 2 \to 3\to 1$. In fact, it is a cone over this path, and $5$ is the apex. I don't think it's a sphere with two discs removed: the link of $5$ is homeomorphic $S^1 \sqcup \mathrm\{pt\}$ (if this is to be a subspace of a 2manifold, the link must be homeomorphic to a subspace of $S^1$). You are correct about the other edges: to other readers, these come from the three triangles we've added. $\endgroup$ Oct 16, 2020 at 8:40

$\begingroup$ @GevaYashfe Thanks, I think I see now. $X$ is more like a sphere with a disc removed (the "missing triangle" $\{1, 2 , 3\}$) and a loop added (the two edges $\{3 , 4 \},\{4, 5\}$), which comes out homotopy equivalent to $S^1$ but in a different way. (My mistake before was that I was drawing the triangle $\{2,3,5\}$ as though the $\{3,5\}$ edge were the nonexistent "composite edge" $\{3,4,5\}$  too much category theory for me!) By the way, why do you say "add an edge $\{1,3\}$" (twice!) at the beginning? That edge comes anyway when the triangle $\{1,3,5\}$ is added. $\endgroup$– Tim Campion ♦Oct 16, 2020 at 20:26

$\begingroup$ @TimCampion That makes sense. I think about simplicial complexes a lot, but my first attempt at an answer was mistaken as well. Thanks for spotting the redundant $\{1,3\}$s. The first one was probably a mistake I made while rephrasing part of the answer. But I didn't notice that the second is redundant as well! I guess I was thinking about it in two steps: first add the "interesting" cycle $1\to 2\to 3$ to $A$, and then add the disc to fill it. I think I'll leave the second $\{1,3\}$ as it is. $\endgroup$ Oct 17, 2020 at 14:00
Take $X$ to be a circle with three 0simplices and three 1simplices. Fix a point, then the union of all contractible subcomplexes containing that point is the whole circle.
EDIT : taking into account the fact that $A$ needs to contain all vertices.
I think it is true. At some point the proof isn't totally formal but I think it can be made rigorous. Let $Y$ be the union of the subcomplexes $C$ with a trivial inclusion in fundamental group and containing $A$, and take a map $f:S^1\to Y$. By some strong version of simplicial approximation you can suppose that there is a simplicial structure on $S^1$ such that $f$ is homotopic to a simplicial inclusion. Take your chosen basepoint of $S^1$, $*$, and run over the circle clockwise. The images of $f$ begin in some subcomplex $C$ of the union and you eventually quit it. Consider first point of the circle after which you quit $C$, say $p$. Then $f(*)$ and $f(p)$ can be linked by a path in $A$ because it contains all the points and is contractible. This path is also in $C$. Glue this path and the path drawn by $f$ from $*$ to $p$, then it makes a loop on $C$ which is trivial in the fundamental group of the whole space $X$. So, $f$ is homotopic in $X$ to the same loop but where you have replace the path from $f(*)$ to $f(p)$ by your chosen path in $A$. But we rather want a homotopy in $Y$. Nevermind : the homotopy can be simplicial approximated in the 2skeleton of $X$ and you can add to $C$ the 2cells needed for the homotopy. This part is less formal but I think it can be written well. Do that by induction until you run out of vertices of $S^1$, and then the new $f$ has values in $A$ so is nulhomotopic.

1$\begingroup$ But their intersection is empty. I required that all complexes in the union contain a given contractible subcomplex $A$. I also forgot to mention that the subcomplex $A$ is on the same set of vertices as $X$. I have now added that to the question. $\endgroup$– erybOct 15, 2020 at 19:09

$\begingroup$ Ok my bad, then take a cell structure on the circle with 3 points and 3 lines. Fix a point, then the union of the contractible simplices containing that point is the whole circle. I'm correcting that in the main answer. $\endgroup$– elidiotOct 15, 2020 at 19:13

$\begingroup$ You are correct, but I forgot to mention that the subcomplex $A$ should have the same set of vertices as $X$. I have now added that to the main question. $\endgroup$– erybOct 15, 2020 at 19:16

1$\begingroup$ If you take a tetrahedron with the standard cell decomposition, and take $A$ to be a point and all 1simplices touching the point, I think it makes a counterexample $\endgroup$– elidiotOct 15, 2020 at 19:25

$\begingroup$ But the tetrahedron has trivial fundamental group, so it would be part of the union? $\endgroup$– erybOct 15, 2020 at 19:33