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A triangulation of a geometric object $G$ is an abstract simplicial complex whose gemoetric realization is homeomorphic to $G$. For example, the abstract complex with facets {A,B},{B,C},{C,A} is a triangulation of the circle $S^1$:

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and the abstract complex with facet {A,B,C} is a triangulation of the disc $B^2$:

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Let $T_2$ be some triangulation of $B^2$, and let $T_1\subseteq T_2$ be a triangulation of $S^1$. Is it possible that there is a simplicial map from $T_2$ to $T_1$, that maps each vertex of $T_1$ to itself?

Here are some examples in which no such map exists.

1. Let $T_1$ and $T_2$ be the triangulations shown above. The only map that maps each vertex of $T_1$ to itself is the identity, and it is not a simplicial map, since it maps {A,B,C} to {A,B,C}, which is not a simplex of $T_1$.

On the other hand, the map $A\to B, B\to B, C\to C$ is simplicial (it maps {A,B,C} to {B,C}, which is a simplex of $T_1$), but it does not satisfy the requirement of mapping each vertex of $T_1$ to itself.

2. Let $T_2$ be the complex with facets {A,B,D},{B,C,D},{C,A,D}; it is a triangulation of $B^2$:

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Let $T_1$ be the complex with facets {A,B},{B,C},{C,A} shown above.

Here, there are three maps that map each vertex of $T_1$ to itself; they map D to either A or B or C. All these candidates are not simplicial. For example, if D is mapped to A, then the simplex {D,B,C} in $T_2$ is mapped to {A,B,C}, which is not a simplex in $T_1$.

Is there a theorem saying that there cannot exist a map satisfying both requirements (simplicial, and maps vertices of $T_1$ to themselves)?

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2 Answers 2

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We have the inclusion simplicial map $\iota : T_1 \to T_2$. Any simplicial map $r : T_2 \to T_1$ that fixes the vertices of $\iota(T_1)$, must also fix the $1$-simplices of $\iota(T_1)$. In other words, $r \circ \iota = Id_{T_1}$.

A simplicial map induces continuous maps in the geometric realization. Then, abusing notation, we have a continuous map $\iota : S^1 \to B^2$ and $r : B^2 \to S^1$ satisfying $r \circ \iota = Id_{S^1}$. Such a map is known as a retraction. Turns out no such retraction of $B^2$ onto $S^1$ can exist, which is an easy consequence of the fact that the fundamental group of $S^1$ is $\pi_1(S^1) = \mathbb{Z}$, whereas $\pi_1(B^2) = 0$.

Thus, no such simplicial map $r : T_2 \to T_1$ can exist in the first place.

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  • $\begingroup$ ". Such a map is known as a retraction" - you mean that $r$ is a retraction, right? $\endgroup$
    – Erel Segal-Halevi
    Feb 12 at 12:15
  • $\begingroup$ Yes, $r$ is a retraction on to the subspace $Im \iota$. The subspace is called a retract of the ambient space. $\endgroup$
    – ChesterX
    Feb 12 at 13:23
  • $\begingroup$ According to Wikipedia, a deformation retraction is a homotopy equivalence, so there cannot be a deformation retraction between spaces with different fundamental groups. But, here you talk about a retract, which is not the same as a deformation retraction. What theorem says that a retract must preserve the fundamental group? $\endgroup$
    – Erel Segal-Halevi
    Feb 13 at 0:57
  • $\begingroup$ Deformation retration (which is a special case of homotopy equivalence) is a much stronger notion than retraction. If $\iota : A \to X$ is the inclusion of a subspace, and $r : X \to A$ is a retraction, i.e, satisfies $r\circ\iota = 1_A$, then we can conclude that that $\iota$ induces a monomorphism at the homotopy level : $1_{\pi_1(A)} = \pi_1(1_A) = \pi_1(r \circ \iota) = \pi_1(r) \circ \pi_1(\iota) \Rightarrow \pi_1(\iota) \text{ is injective}$. In other words, $\pi_1(A)$ must be a subgroup of $\pi_1(X)$. In your situation, $\pi_1(S^1) = \mathbb{Z}$ cannot be a subgroup of $\pi_1(B^2) = 0$. $\endgroup$
    – ChesterX
    Feb 13 at 2:45
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Here is a proof, using Sperner's lemma, if $T_1$ is the standard triangle. This is not an essential restriction but reducing from the general case is not completely straightforward, especially in higher dimensions. Order the vertices of the triangle (say $A < B < C$ for concreteness).

Let $f$ be such a map and mark each vertex $v$ of $T_2$ with the smallest vertex in the support of $f(v)$ (so if $f(v)$ is on the edge $AB$ or $AC$ but not the vertices $B$ or $C$ then mark it with $A$, if it's on the edge $BC$ but not $C$ then mark it with $B$, otherwise mark it with $C$). Since $f$ is identity on $T_1$, it is easy to check that any triangle and its image under $f$ has the same markings. But every simplex of $T_1$ is missing some vertex in its support so it cannot be rainbow, whereas Sperner's lemma implies that some simplex of $T_2$ is rainbow.

It should be easy to see how this generalizes to higher dimensions.

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