# Can there be a simplicial map from a triangulation of the disc to a triangulation of the circle?

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 $$S^1$$: and the abstract complex with facet {A,B,C} is a triangulation of the disc $$B^2$$: 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 $$B^2$$: 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)?

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.

• ". Such a map is known as a retraction" - you mean that $r$ is a retraction, right? Feb 12 at 12:15
• Yes, $r$ is a retraction on to the subspace $Im \iota$. The subspace is called a retract of the ambient space. Feb 12 at 13:23
• 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? Feb 13 at 0:57
• 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 $\pi_1(B^2) = 0$. Feb 13 at 2:45

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.