# 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 $\pi_1$?

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?

• 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 $\bigcup C$. Oct 15, 2020 at 22:06
• 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 $A$. Oct 15, 2020 at 22:27
• (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.) Oct 16, 2020 at 20:34
• 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 $\tilde{X}$.
– eryb
Oct 17, 2020 at 8:04

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 $$\bigcup C$$.

• 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. Oct 15, 2020 at 22:03
• @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 2-manifold, 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. Oct 16, 2020 at 8:40
• @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. Oct 16, 2020 at 20:26
• @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. Oct 17, 2020 at 14:00

Take $$X$$ to be a circle with three 0-simplices and three 1-simplices. 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 2-skeleton of $$X$$ and you can add to $$C$$ the 2-cells 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.

• 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.
– eryb
Oct 15, 2020 at 19:09
• 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. Oct 15, 2020 at 19:13
• 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.
– eryb
Oct 15, 2020 at 19:16
• If you take a tetrahedron with the standard cell decomposition, and take $A$ to be a point and all 1-simplices touching the point, I think it makes a counter-example Oct 15, 2020 at 19:25
• But the tetrahedron has trivial fundamental group, so it would be part of the union?
– eryb
Oct 15, 2020 at 19:33