# Higher homotopy groups in terms of the fundamental groupoid

Let $X$ be a topological space. Then we can construct the following structure. Let an $n$-morphism be a map $I^n\to X$. We can view $n+1$ morphisms exactly as homotopies between $n$-morphisms.

Let $f,g$ and $h$ be $n$-morphisms and $H$ and $G$ be homotopies from $f$ to $g$ and from $g$ to $h$ respectively. Then we define $G\circ_{n+1}H$ to be the homotopy from $f$ to $h$ defined by

$$G\circ_{n+1}H(x_1,\dots,x_n,x)= \begin{cases} H(x_1\dots,x_n,2x)\;\;\;\;\;\,\text{ if x\in\left[0,\frac{1}{2}\right]}\\ G(x_1\dots,x_n,2x-1)\text{ if x\in\left[\frac{1}{2},1\right]} \end{cases}$$

It is easy to see that $\circ_n$ is associative and for every $n$-morphism f, there is the trivial homotopy with itself $\text{id}_f$. So we see that for every $n\geq0$, we have a category with objects as $n$-morphisms and arrows as $n+1$-morphisms.

Then we can define the map $\bar{H}\colon I^{n+1}\to X$ as $\bar{H}(x_1\dots,x_n,x)=H(x_1\dots,x_n,1-x)$. Then $\bar{H}$ is a homotopy from $g$ to $f$. Then composite $\bar H\circ_{n+1}H$ homotopy fromm $f$ to itself is defined by

$$\bar H\circ_{n+1}H(x_1,\dots,x_n,x)= \begin{cases} H(x_1\dots,x_n,2x)\;\;\;\;\;\,\text{ if x\in\left[0,\frac{1}{2}\right]}\\ H(x_1\dots,x_n,2-2x)\text{ if x\in\left[\frac{1}{2},1\right]} \end{cases}$$ Define $\alpha$ to be the homotopy ($n+2$-morphism) from $\text{id}_f$ to $\bar H\circ_{n+1}H$ defined by

$$\alpha(x_1,\dots,x_n,x,t)= \begin{cases} H(x_1,\dots,x_n,2tx)\;\;\;\;\;\;\;\;\;\,\text{if x\in\left[0,\frac{1}{2}\right]}\\ H(x_1,\dots,x_n,2t(1-x))\;\text{if x\in\left[\frac{1}{2},1\right]} \end{cases}$$ So we see that for every $n+1$-morphism, there is 'weak' inverse. So This is in some sense, a weak $\infty$-groupoid.

Now, we choose a point $x_0$. This is a $0$-morphism. So, we can look at $\text{Hom}(x_0,x_0)$ (set of $1$-morphisms from $x_0$ to $x_0$). Now, we can formally turn all the higher weak equivalences into equalities (by quotienting by this hom-set by the appropriate relation). Then, the group left over is nothing but $\pi_1(X,x_0)$.

My questions then are

• We can perform an equivalent process at higher levels. That is, choose an $n$-morphism and turn all the $n+2$-morphisms into strong equivalences. Then we have the group $\text{Hom}_{n+1}(f,f)$ ($n+1$ automorphisms of $f$ identified upto equivalence). Are these groups classically interesting? Do they have good properties (under homotopy equivalnce, for example)?
• Can we describe higher homotopy groups in terms of this weak $\infty$-groupoid?
• See e.g. the book "Algebraic Topology" by Allen Hatcher (math.cornell.edu/~hatcher/AT/AT.pdf) p. 340. You must consider "n-morphisms" which map the boundary $\partial I^n$ to a fixed basepoint $x_0 \in X$. – Paul Frost Apr 30 '18 at 10:18
• @Paul, I get that. But, are the $\pi_n$ just the automorphism groups at higher levels? I don't think so as my construction involves the choice of an $n$-morphism, not a basepoint. My other question was to define the $\pi_n$ from the groupoid, without referencing the underlying space. I already know the classical definition (given in Hatcher). – Chetan Vuppulury Apr 30 '18 at 10:22
• I could imagine that the groups $Hom_n(f,f)$ contain some interesting information although I have not seen them in the literature (perhaps someone else has?).They are generalizations of the classical homotopy groups, but I do not see how the latter could be recovered from your weak $\infty$-groupoid. It seems to me that one is not able to identify the elements corresponding to $\pi_n(X,x_0)$ if one only uses the abstract groupoid-structure. – Paul Frost Apr 30 '18 at 10:45
• One more remark: $\circ_{n+1}$ does not produce a category (it is not associative and $id_f$ is not an identity), You have to define arrows as equivalence classes of $(n+1)$-morphisms. It seems the only reasonable way to define $(n+1)$-morphisms $H, K$ from $f$ to $g$ to be equivalent is to require the existence of an $(n+2)$-morphism $\Sigma$ from $H$ to $K$ such that $\Sigma(x,0,t) = f(x) = H(x,0) = K(x,0)$ and $\Sigma(x,1,t) = g(x) = H(x,1) = K(x,1)$ for all $t$.This makes all arrows isomorphisms. – Paul Frost Apr 30 '18 at 14:47
• @PaulFrost there's no need for strict associativity in a weak $\infty$-groupoid. – Kevin Carlson May 1 '18 at 2:24

## 3 Answers

From your definition, it's really not clear what groups you're referring to. That's because your weak groupoid is cubical, in that an $n$-morphism is shaped like an $n$-dimensional hypercube. In particular, it has $n$ $(n-1)$-dimensional domains and $n$ codomains, corresponding to the $2n$ top-dimensional faces of the hypercube.

So, when you claim to reconstruct $\pi_1(X,x)$, what exactly is the relation you're putting on $Hom(x,x)$? You say you're making all higher morphisms into equalities, which suggests to me the relation $f\sim g$ if there exists a 2-morphism with $f$ and $g$ as two opposing faces. This is the relation that $f$ and $g$ are homotopic without endpoints fixed. The resulting quotient admits no natural group structure-it's the conjugacy classes in $\pi_1(X,x)$.

The relation that will produce groups is: $f\sim g$ if there exists a 2-morphism with $f$ and $g$ as two opposing faces and in which the other two faces are degenerate. That is, the relation is precisely that there exists a homotopy between $f$ and $g$ with endpoints fixed, which is precisely the defining relation for $\pi_1(X,x)$.

In higher dimensions, the appropriate relation is analogous: two $n$-morphisms $f,g$ with boundaries degenerate at $x$ are equivalent if they form opposite faces of an $n+1$-morphism in which all other faces are degenerate. Again, this is precisely the definition of $\pi_n(X,x)$.

What about automorphism groups of morphisms? For instance, given a 1-morphism $f:x\to y$, we have a group of all 2-morphisms $H$ whose boundary paths are $f,x,f,$ and $y$, where $x$ and $y$ stand for their constant paths, taken up to 3-morphisms with $H$ and $K$ as two faces and $x,f,y,$ and $f$ as the other four. This is nothing more or less than $\pi_1(P(x,y),f)$, where $P(x,y)$ denotes the space of paths in $X$ between $x$ and $y$. This may look like an unfamiliar space, but composing with $f^{-1}$ gives a homotopy equivalence with $\Omega(X,x)$, the space of loops based at $x$, which is well known to have $\pi_2 X$ for its fundamental groups.

There are still more groups to be found in your groupoid! For instance, given a loop $\gamma$ in $X$, an appropriate relation on certain 2-morphisms will reproduce $\pi_1(LX,\gamma)$, which is hard to express precisely in terms of $X$ if $\gamma$ isn't contractible. In fact, you can do quite a lot more: your groupoid contains enough information to reconstruct $X$ up to homotopy equivalence (assuming it's of the homotopy type of a CW complex) so that in principle it contains every homotopy meaningful group that can be constructed out of $X$.

However, there are some strict limits on discovering new such groups as you wanted to. By Heller's representability theorem, any functor on based spaces into groups which respects homotopies, products, and relative path spaces must be induced by homotopy classes of maps out of some space. That's why we shouldn't be surprised to find a homotopy group of a space related to $X$ above-it's almost impossible to do anything else.

• Thanks for the great answer! Do you know any references which approach homotopy theory through this kind of a structure? – Chetan Vuppulury May 1 '18 at 10:05
• @ChetanVuppulury Yes and no. It's standard by now to approach homotopy theory via Kan complexes, which can be thought of as $\infty$-groupoids in which the composition itself is only well defined up to homotopy. Simplicial Homotopy Theory by Goerss-Jardine is great, though it won't talk in terms of higher higher groupoids. The idea that Kan complexes are equivalent to higher groupoids is an incarnation of the homotopy hypothesis, one of the fundamental problems of higher category theory. But there's not yet a convincing algebraic development of the homotopy theory of $\infty$-groupoids. – Kevin Carlson May 1 '18 at 14:03

The groups you define are merely isomorphic to the usual higher homotopy groups.

Let me do the case $n = 1$ otherwise it becomes notationally tedious. Fix some $f : [0,1] \to X$. You're looking at the set $$\{ H : [0,1]^2 \to X \mid H(0,t) = H(1,t) = f(t) \} / \sim$$ where $\sim$ is the relation "homotopic", and a group structure given by concatenation. I claim that this is simply $\pi_2(X, f(0))$. I'll write maps in one directions, and it's easy that it's compatible with the group structure and bijective (by building a similar map in the reverse direction).

Start with a representative of $\pi_2(X,f(0))$, say $A : [0,1]^2 \to X$ such that $A(\partial[0,1]^2) = \{f(0)\}$. You can build a $H : [0,1]^2 \to X$ as above by: $$H(s,t) = \begin{cases} f(t(1-3s)), & \text{if } 0 \le s \le 1/3; \\ A(3s-1,t), & \text{if } 1/3 \le s \le 2/3; \\ f(t(3s-2)), & \text{if } 2/3 \le s \le 1. \end{cases}$$ Basically for $0 \le s \le 1/3$ we build a homotopy from $f$ to the constant map equal to $f(0)$; for $1/3 \le s \le 2/3$ we use $A$; and for $2/3 \le s \le 1$, we again build a homotopy from $t \mapsto f(0)$ to $f$.

• In fact the groups $Hom_{n+1}(f), Hom_{n+1}(g)$ are isomorphic if $f, g$ are homotopic. You claim that if $c : I^n \to X$ is a constant map with value $x_0$, then $Hom_{n+1}(c) \approx \pi_{n+1}(X,x_0)$. You certainly have a homomorphism $\pi_{n+1}(X,x_0) \to Hom_{n+1}(c)$, but why is it bijective? In $\pi_{n+1}(X,x_0)$ all elements are represented by $A : I^{n+1} \to X$ such that $A(\partial I^{n+1}) = \lbrace x_0 \rbrace$, but in $Hom_{n+1}(c)$ you only have $A(I^n \times \lbrace 0, 1 \rbrace) = \lbrace x_0 \rbrace$. – Paul Frost Apr 30 '18 at 15:14

The groups $Hom_{n+1}(f), Hom_{n+1}(g)$ are isomorphic if $f, g$ are homotopic. Therefore it suffices to consider the group $Hom_{n+1}(c)$ where $c : I^n \to X$ is a constant map with value $x_0$. Define

$\varphi : \pi_1(X,x_0) \to Hom_{n+1}(c), \varphi([\omega]) = [\overline{\omega}]$ with $\overline{\omega}(x,t) = \omega(t)$.

This is clearly a group homomorphism. We show that it is an isomorphism.

$\varphi$ is surjective: Let $[A] \in Hom_{n+1}(c)$. Define

$\omega_A : I \to X, \omega_A(t) = A(0,t)$ with $0 = (0,...,0) \in I^n$,

$\Sigma : I^{n+1} \times I \to X, \Sigma(x,s,t) = A(t \cdot x, s)$.

Then $\Sigma_0 = \overline{\omega_A}$, $\Sigma_1 = A$. This shows $\varphi([\omega_A]) = [A]$.

That $\varphi$ is injectice can be shown similarly.

• Why are we only considering $\text{Hom}_n(f,f)$ for $f$ being the constant map? Do other $f$s contain more information? – Chetan Vuppulury May 1 '18 at 4:11
• @ChetanVuppulury No, they do not. Each $f$ is homotopic to a constant map since the domain $I^n$ is contractible. – Paul Frost May 1 '18 at 7:48
• According to Kevin Carlson's comment above my answer is possibly not adequate since it is based on a different interpretation of $Hom_{n+1}(f,f)$. – Paul Frost May 1 '18 at 8:02
• @PaulFrost Based on your interpretation in the top-level comments, your version of these sets does not admit a group operation, just as I explained in the 1-dimensional case in my answer. – Kevin Carlson May 1 '18 at 14:07
• @KevinCarlson As answers and comments show, Chetan Vuppulury's question is inspriring. Although it is not perfectly precise, I think that you - unlike me - captured the spirit of the question and I upvoted your answer. Let me nevertheless clarify my interpretation. For any two maps $f,g : X \to Y$ let $\mathfrak{H}(f,g)$ be the set of homotopies $H : f \simeq g$. Define $H \sim H'$ if there exists a homotopy $\Sigma : X \times I \times I \to Y$ rel. $X \times \lbrace 0, 1\rbrace$ from $H$ to $H'$. Let $Hom(f,g) = \mathfrak{H}(f,g)/\sim$. This produces a grouppoid whose objects – Paul Frost May 1 '18 at 23:07