I learn from Gelfand and Manin's Methods of Homological Algebra, Exercise 2 for I.4 that two maps $f,g\colon X\to Y$ between simplicial sets $X,Y$ are simply homotopic (maybe usually called simplicial homotopic, as Zhen Lin pointed out) if and only if there exists $h\colon\Delta[1]\times X\to Y$ such that either $f=h\circ p_0,g=h\circ p_1$ or $f=h\circ p_1,g=h\circ p_0$ where the simplicial set $\Delta[1]\times X$ is the diagonal of the bisimplicial set $\Delta[1]\times X$ (sorry for abuse of notation), and $p_0,p_1\colon\Delta[1]\times X\to X$ are projections to vertices $0$ and $1$, and $f,g\colon X\to Y$ are homotopic if there exists a chain $f=f_0,f_1,\dotsc,f_{n+1}=g$ such that $f_j,f_{j+1}$ are simply homotopic for any $0\le j\le n$.

Meanwhile, when we're considering simplicial maps $f,g\colon K\to L$ between simplicial complexes $K,L$, we define $f,g$ are contiguous if, for each simplex $\sigma\in K$, $f(\sigma),g(\sigma)$ lies in a common simplex in $L$, and $f,g$ are contiguous equivalent if there exists $f=f_0,f_1,\dotsc,f_{n+1}=g$ such that $f_j,f_{j+1}$ are contiguous for any $0\le j\le n$. The definition is taken from Singer and Thorpe's Lecture Notes on Elementary Topology and Geometry.

My question arises: since simplicial complexes have simplicial structures, namely could be naturally associated with simplicial sets, does these concepts coincide? Namely, is it true that $f,g$ are homotopic if and only if they are contiguous equivalent in the context of simplicial complexes?

One should note that the concept of simply homotopy relies, informally, speaking, on the canonical triangulation of the $\Delta[1]\times X$ as a bisimplicial set, but the one of contiguous doesn't.

Any help? Thanks!

  • $\begingroup$ Contiguous is equivalent to homotopic in the usual sense provided that you give yourself the freedom to barycentrically divide $K$ as necessary by simplicial approximation. Simple homotopy (at least if this agrees with what usually goes by this name) is strictly stronger than homotopic in the usual sense; there is an obstruction coming from Whitehead torsion, see en.wikipedia.org/wiki/Simple-homotopy_equivalence. $\endgroup$ – Qiaochu Yuan Sep 11 '14 at 8:05
  • $\begingroup$ @QiaochuYuan Simply homotopy of $f,g\colon X\to Y$ here, informally speaking, is just a simplicial map $F\colon\Delta[1]\times X\to Y$ from a cylinder $\Delta[1]\times X$ (triangulated canonically, where $\Delta[1]$ is just the simplicial set associated with $I=[0,1]$, i.e., $1$-simplex) to $Y$, both of which are simplicial sets. Sorry for my ignorance. I don't know what you're referring to. Thanks, anyway. $\endgroup$ – Yai0Phah Sep 11 '14 at 9:12
  • 2
    $\begingroup$ I believe this is usually called "simplicially homotopic". $\endgroup$ – Zhen Lin Sep 11 '14 at 9:21
  • $\begingroup$ @ZhenLin It seems that you are right, if the definition of the product is essentially same as the one in Goerss & Jardine. It might be a mistranlation from Russian. $\endgroup$ – Yai0Phah Sep 11 '14 at 16:10

CONTIGUOUS implies Topologically Homotopic (finding a family of curves, like you wrote, between the two maps) Topological Homotopy of continuos maps between simplicial spaces implies Homotopic Equivalence (in the chain complex), a bit harder to prove (using Simplicial Approximations and Baricentric Operator)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.