Simplicial homotopy

In algebraic topology, a simplicial homotopy is an analog of a homotopy between topological spaces for simplicial sets. Precisely,^{pg 23} if

$f,g:X\to Y$

are maps between simplicial sets, a simplicial homotopy from f to g is a map

$h:X\times {\mathrm{\Delta }}^1\to Y$

such that the restriction of $h$ along $X\simeq X\times {\mathrm{\Delta }}^0\stackrel{0}{\hookrightarrow }X\times {\mathrm{\Delta }}^1$ is $f$ and the restriction along $1$ is $g$; see [1]. In particular, $f(x)=h(x,0)$ and $g(x)=h(x,1)$ for all x in X.

Using the adjunction

$\mathrm{Hom}(X\times {\mathrm{\Delta }}^1,Y)=\mathrm{Hom}({\mathrm{\Delta }}^1\times X,Y)=\mathrm{Hom}({\mathrm{\Delta }}^1,\underset{\_}{\mathrm{Hom}}(X,Y))$,

the simplicial homotopy $h$ can also be thought of as a path in the simplicial set $\underset{\_}{\mathrm{Hom}}(X,Y).$

A simplicial homotopy is in general not an equivalence relation. However, if $\underset{\_}{\mathrm{Hom}}(X,Y)$ is a Kan complex (e.g., if $Y$ is a Kan complex), then a homotopy from $f:X\to Y$ to $g:X\to Y$ is an equivalence relation. Indeed, a Kan complex is an ∞-groupoid; i.e., every morphism (path) is invertible. Thus, if h is a homotopy from f to g, then the inverse of h is a homotopy from g to f, establishing that the relation is symmetric. The transitivity holds since a composition is possible.

Simplicial homotopy equivalence

If $X$ is a simplicial set and $K$ a Kan complex, then we form the quotient

$[X,K]=\mathrm{Hom}(X,K)/\sim$

where $f\sim g$ means $f,g$ are homotopic to each other. It is the set of the simplicial homotopy classes of maps from $X$ to $K$. More generally, Quillen defines homotopy classes using the equivalence relation generated by the homotopy relation.

A map $K\to L$ between Kan complexes is then called a simplicial homotopy equivalence if the homotopy class $[f]$ of it is bijective; i.e., there is some $g$ such that $fg\sim {\mathrm{id}}_L$ and $gf\sim {\mathrm{id}}_K$.

An obvious pointed version of the above consideration also holds.

Simplicial homotopy group

Let $S^1$ be the pushout ${\mathrm{\Delta }}^1{\bigsqcup }_{\partial {\mathrm{\Delta }}^1}1$ along the boundary $S^0=\partial {\mathrm{\Delta }}^1$ and $S^n=S^1\wedge \cdots \wedge S^1$ n-times. Then, as in usual algebraic topology, we define

${\pi }_{n}X=[S^n,X]$

for each pointed Kan complex X and an integer $n\ge 0$. It is the n-th simplicial homotopy group of X (or the set for $n=0$). For example, each class in ${\pi }_{0}X$ amounts to a path-connected component of $X$.

If $X$ is a pointed Kan complex, then the mapping space

$\mathrm{\Omega }X={\mathrm{Map}}_X(x_0,x_0)$

from the base point to itself is also a Kan complex called the loop space of $X$. It is also pointed with the base point the identity and so we can iterate: ${\mathrm{\Omega }}^{n}X$. It can be shown

${\mathrm{\Omega }}^{n}X=\underset{\_}{\mathrm{Hom}}(S^n,X)$

as pointed Kan complexes. Thus,

${\pi }_{n}X={\pi }_0{\mathrm{\Omega }}^{n}X.$

Now, we have the identification ${\pi }_0{\mathrm{Map}}_C(x,y)={\mathrm{Hom}}_{\tau (C)}(x,y)$ for the homotopy category $\tau (C)$ of an ∞-category C and an endomorphism group is a group. So, ${\pi }_{n}X$ is a group for $n\ge 1$. By the Eckmann-Hilton argument, ${\pi }_{n}X$ is abelian for $n\ge 2$.

An analog of Whitehead's theorem holds: a map $f$ between Kan complexes is a homotopy equivalence if and only if for each choice of base points and each integer $n\ge 0$, ${\pi }_n(f)$ is bijective.

See also

• Kan complex
• Dold–Kan correspondence (under which a chain homotopy corresponds to a simplicial homotopy)
• Simplicial homology
• Homotopy category of an ∞-category

References

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External links

• Simplicial homotopy

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