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If someone asked me the question for the fundamental group, I would answer as follows:

  1. The connection to classification of covering spaces.
  2. The fundamental group of many spaces is an object of independent interest. For instance, for an Elliptic curve over the complex numbers, the fundamental group is the lattice defining the curve or equivalently, it is related to the torsion points on the curve.
  3. Related to 1, the arithmetic fundamental group is closely related and the arithmetic fundamental group is itself very important. For instance, the Galois groups of fields is an example.
  4. The fundamental group offers very natural proofs of fundamental theorems like the fundamental theorem of algebra.

However, for the higher homotopy groups, the best answer I could give would be something along the lines of the the long exact sequence of homotopy groups]1 for fibrations. Maybe the Hurewicz theorem is also an answer to my question except that I think the Hurewicz theorem is usually used to get information about the homotopy groups from the homology groups. If this is not true, that would be an interesting answer too.

I am almost sure this is entirely due to my background (in arithmetic geometry) and lack of formal training in algebraic topology and that the higher homotopy groups are indeed a natural object to study.

Ideally, I would appreciate answers that either connect the higher homotopy groups to important invariants of spaces that were already studied (1,2, 3 above) or proofs of statements not about the higher homotopy groups that however use the higher homotopy groups in an essential way (4 above and I guess the long exact sequence comes under here).

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    $\begingroup$ The OP requested CW status (there being already a 'big list' of good answers). $\endgroup$
    – Todd Trimble
    Oct 21, 2017 at 23:45

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Some good answers have already been given. To my mind, though: a really obvious thing one wants to do in topology is to classify topological spaces, or more reasonably, to classify CW-complexes, at least up to homotopy equivalence.

  • You build CW-complexes by attaching cells.

  • To attach a cell, you just have to specify the attaching map, i.e., the continuous map from the boundary of the cell to the skeleton you've already constructed.

  • The boundary of a cell is a sphere.

  • The homotopy type of the space you get by attaching a cell depends only on the homotopy class of the attaching map.

So homotopy groups are telling you the ways you can attach a cell. So the question of how many homotopy types of CW-complexes you can build with some given property typically comes down to a computational problem about homotopy groups.

For example: it sounds like you are an arithmetic geometer who already is convinced of the utility of homology groups and the fundamental groups, so suppose you are doing your work in arithmetic geometry, and you find yourself confronted with two different smooth schemes over Spec Z whose underlying analytic spaces are simply-connected and have homology groups isomorphic to Z in degrees 0, 8, and 14, and trivial in all other degrees. You knock on the door of the friendly topologist down the hall and ask the topologist whether your two analytic spaces are necessarily homotopy-equivalent.

  • The topologist first checks with you to make sure there's some general theorem which ensures that these spaces are homotopy-equivalent to CW-complexes, and then observes that a minimal CW-decomposition of any such space ought to have a 0-cell, an 8-cell, and an 14-cell, since anything else would give you the wrong homology.
  • The 8-cell has to be attached trivially to the 0-cell for silly reasons, so the 8-skeleton of any such CW-complex must be S^8.
  • Then the topologist points out that the attaching map for the 14-cell must be a map from its boundary, a 13-sphere, to an 8-sphere.
  • The homotopy group $\pi_{13}(S^8)$ is in the stable range, by the Freudenthal suspension theorem, so the topologist shows you a 2-primary Adams spectral sequence chart and points to the empty 5-column, and says "So the 5-stem is 2-locally trivial."
  • Then the topologist tells you a bit (maybe more than you wanted to hear) about how the alpha family and beta_1 work at odd primes, ending with the conclusion that the 5-stem (the fifth stable homotopy group of spheres) also vanishes at all odd primes, and hence $\pi_{13}(S^8)$ is trivial.
  • Consequently there is only one homotopy class of attaching map for that 14-cell which has been attached to S^8. So your two analytic spaces are homotopy-equivalent.

If you have more than just two cells in positive dimensions, Toda brackets are a convenient way to organize the algebra of the attaching maps, in order to reduce these kinds of classification problems to algebraic problems in the homotopy groups of spheres.

Never used this site before--hope I didn't write anything too critically stupid. Sorry if I did.

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I think that obstruction theory is one of the most important reasons to study homotopy groups. If you are interested in studying the possible homotopy classes of maps $X \to Y$ of spaces where $X$ has the homotopy type of a CW-complex, obstruction theory gives you an method for doing so by building it iteratively on skeleta. You get obstructions in the cohomology groups $$ H^{s+1}(X; \pi_s(Y)) $$ which tell you whether your partial construction can be extended, and you get difference classes in the cohomology groups $$ H^s(X;\pi_s(Y)) $$ which tell you how many fundamentally different extensions you can get. (I'm simplifying slightly by assuming $Y$ is simply-connected.) There are also relative versions of this which allow you to assume that your map is already defined on a subspace of $X$ or that you are trying to make this map compatible with maps from $X$ and $Y$ to a fixed space $Z$.

I like to think about it this way. The (co)homology of $X$ typically tells us information about how $X$ is built using basic cells $S^n$ and $D^{n+1}$, while the homotopy of $Y$ tells us the possible ways that those basic cells can be mapped in. The two are natural counterpoints to each other once we've settled on "cells" as our building blocks.

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    $\begingroup$ And note that in the special case $Y = BG$ we are classifying principal $G$-bundles on $X$, which is a very large and important generalization of classifying covering spaces (which correspond to $G = S_n$ for $n$-sheeted covers). $\endgroup$ Oct 21, 2017 at 18:56
  • $\begingroup$ @QiaochuYuan regarding covering spaces, don't you mean $G$ is any transitive subgroup of $S_n$? $\endgroup$ Mar 12, 2018 at 20:29
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The calculation of the ring of manifolds up to cobordism was made possible thanks to identifying this ring with the homotopy groups of certain Thom spaces, and then applying methods of homotopy theory.

The homotopy groups of classifying spaces like $BO$ and $BU$ are very important because of their connection to topological K-theory.

And to this day the only way to even define higher algebraic K-theory groups of a ring is as the homotopy groups of some (infinite loop) space. Of course you could ask about the importance of higher algebraic K-theory, but that's a subject for another thread. And of course, methods of homotopy theory have been extremely useful in calculating algebraic K-theory.

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    $\begingroup$ Why the downvote? $\endgroup$ Oct 21, 2017 at 19:22
  • $\begingroup$ Wouldn't this argue that homotopy groups are the wrong definition, and the right definition is stable homotopy groups? (To be clear, I'm not the downvoter.) $\endgroup$ Oct 23, 2017 at 0:29
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    $\begingroup$ It is true all my examples are homotopy groups of infinite loop spaces, but they are not stable homotopy groups of spaces. From this point of view, stable homotopy groups are a special case of unstable ones, rather than a different invariant. So I would rather say that my answer leaves open the question of motivating the study of homotopy groups of non-infinite loop spaces. Several of the other excellent answers fill this gap in various ways. $\endgroup$ Oct 23, 2017 at 7:26
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I think there's something fundamental missing from all the other answers so far: the modern realization that topological spaces are distinct from $\infty$-groupoids.

Suppose you didn't know about higher homotopy groups at all. In fact, suppose you didn't even know about the fundamental group, and maybe you didn't even know about the notion of homotopy! Here's a way they could have been invented. Many of the other answers touch on parts of this story, but I think the whole thing together gives us additional insight.

You may or may not be familiar with the notion of $\infty$-groupoid. Technically it's complicated, but intuitively I think it is (or should be) quite simple: it's just about taking seriously the idea that any "collection of things" can, and often should, come with a notion of "isomorphism" or "sameness" between any two of those things -- and that those isomorphisms are again a "collection", hence can come with their own notion of isomorphism, ad infinitum. In more algebro-geometric language, an $\infty$-groupoid is the natural sort of "thing" for a moduli space to be: you have some things, and you specify when some of those things are the same as each other, and then you specify when some of those "samenesses" are the same as each other, and so on.

Fundamental examples of $\infty$-groupoids arise by repeatedly considering collections of sets and set-like objects. The collection of all sets is (ignoring questions of size, i.e. proper-class-ness or not) best regarded not as a set but as a groupoid, since usually we care not whether two sets (or objects built out of sets like groups, rings, fields, topological spaces, etc.) are "equal" in the sense of having exactly the same elements, but whether they are isomorphic. A groupoid is just an $\infty$-groupoid in which no two "ways that two objects are the same" can themselves be the same in any nontrivial way. But for similar reasons, the collection of groupoids is naturally regarded as a 2-groupoid, which is trivial after the "ways that two samenesses are the same" level; the collection of 2-groupoids is a 3-groupoid; and so on. This doesn't get you all the way to $\infty$-groupoids by induction, but it's natural to want a context in which $n$-groupoids live for all $n$ simultaneously, and moreover at the infinity point it stabilizes: the collection of $\infty$-groupoids is another $\infty$-groupoid.

Now an $\infty$-groupoid is a rather complicated structure, so when working with them it's useful to be able to extract simpler invariants. One obvious useful invariant is the set of isomorphism classes of objects, $\pi_0(G)$ (the "coarse moduli set"). And since the set $G(x,y)$ of isomorphisms between any two objects of an $\infty$-groupoid is again an $\infty$-groupoid, we can also form $\pi_0(G(x,y))$, and also $\pi_0(G(x,y)(f,g))$ and so on. However, if $x$ and $y$ are not isomorphic then $G(x,y)$ is empty, while if they are isomorphic then $G(x,y) \simeq G(x,x)$ by composing with any isomorphism $x\cong y$. Thus, "all the information" is carried by applying this only to automorphisms, giving $\pi_0(G(x,x))$, $\pi_0(G(x,x)(\mathrm{id}_x,\mathrm{id}_x))$, and so on: these are the "homotopy groups" $\pi_1(G,x)$, $\pi_2(G,x)$, etc. The sense in which this is "all the information" is that a map of $\infty$-groupoids $f:G\to H$ is an equivalence if and only if it induces isomorphisms on $\pi_n$ for all $n\ge 0$.

Note that I haven't said anything about "topological spaces" or "weak homotopy equivalence" yet. The last sentence of the previous paragraph is a theorem about objects called "$\infty$-groupoids" that can be motivated and justified on grounds completely independent from topology.

The topology enters the picture by way of a functor $\Pi : \mathrm{Top} \to \infty \mathrm{Gpd}$ called the fundamental $\infty$-groupoid of a topological space. The $\infty$-groupoid $\Pi X$ can be built explicitly by saying "take its points to be points of $X$, its isomorphisms $x\cong y$ to be paths in $X$, its 2-isomorphisms to be homotopies in $X$, and so on." We then define the homotopy groups of $X$ to be those of its fundamental $\infty$-groupoid, $\pi_n(X,x) = \pi_n(\Pi(X),x)$, and define a continous map $f:X\to Y$ to be a weak homotopy equivalence if $\Pi f : \Pi X \to \Pi Y$ is an equivalence. Then we can observe that for a large class of spaces, such as CW complexes, weak homotopy equivalences coincide with homotopy equivalences, and fundamental groups have useful relations to homology, obstruction theory, etc.

However, the construction of $\Pi$ can be motivated in an even more basic way, that doesn't depend on deciding for some other reason that "paths and homotopies" in a space are interesting, or that spaces built by attaching cells are interesting. Note that a bare set can be regarded as an $\infty$-groupoid with no nontrivial isomorphisms, but also as a topological space with the discrete topology. Thus, it's not unreasonable to imagine that there might be a category $\mathcal{T}$ (or more precisely an $\infty$-category) that contains topological spaces and $\infty$-groupoids as full subcategories whose intersection consists of the bare sets. In general, the objects of $\mathcal{T}$ will be sets that have "both topology and $\infty$-groupoid structure". For instance, the groupoid of topological spaces naturally inherits topologies on its spaces of isomorphisms; hence it is a "topological 2-groupoid" which is a particular object of $\mathcal{T}$.

There is more than one way to construct such a category $\mathcal{T}$. However, what they have in common is that while the inclusion $\mathrm{Top} \to \mathcal{T}$ does not preserve all colimits, it does at least preserve unions of open subsets. That is, if a space $X$ is the union of open subspaces $X = \bigcup_i U_i$, then the diagram consisting of all the inclusions $U_i \to X$ is a colimiting cone in $\mathcal{T}$ under the diagram consisting of the spaces $U_i$ and all of their intersections $U_i\cap U_j$, $U_i\cap U_j\cap U_k$, etc. This should seem reasonable if you recall that the very notion of topological space is (or can be) defined in terms of open subsets that are closed under arbitrary unions: we're just saying that the inclusion $\mathrm{Top} \to \mathcal{T}$ respects the "basic building blocks of topology".

Now it turns out that the other inclusion $\infty \mathrm{Gpd} \to \mathcal{T}$ is a reflective subcategory (sub-$\infty$-category), i.e. the inclusion functor has a left adjoint, which I will denote ʃ. What this means is that if $X$ is a topological space and $G$ an $\infty$-groupoid, then morphisms $X\to G$ in $\mathcal{T}$ are equivalent to maps of $\infty$-groupoids ʃ $X \to G$.

Why is this interesting? Well, it's reasonable to view a map $X\to G$ as a "family of objects of $G$ parametrized continuously by $X$". If this needs further motivation, consider that if $X$ and $Y$ are both spaces, then a map $X\to Y$ is certainly a "family of points of $Y$ parametrized continuously by $X$". Now generalize this to the case when $Y$ is an arbitrary object of $\mathcal{T}$. For instance, if $Y$ is the groupoid of topological spaces, regarded as an object of $\mathcal{T}$ as mentioned above, then a map $X\to \mathcal{T}$ is exactly a fiber bundle over $X$: a family of topological spaces (the fibers) "continuously parametrized by $X$". Now specialize to the case when $Y$ lies in $\infty\mathrm{Gpd}$. For instance, if $G$ is the groupoid of sets, then a map $X\to G$ is "a continuously $X$-indexed family of sets", which happens to coincide with a covering space of $X$. Thus, the reflectivity of $\infty \mathrm{Gpd}$ says that the $\infty$-groupoid ʃ $X$ contains all the information about $X$-indexed families of objects of (non-topological) $\infty$-groupoids.

Now it happens -- as a theorem, not a definition -- that this left adjoint reflector ʃ turns out to be the fundamental $\infty$-groupoid! In other words, even if you don't know why to care about paths and homotopies, if you only know to care about topological spaces and $\infty$-groupoids, and want to talk about them in a natural common context that respects the basic building blocks of topology, and to talk about objects of $\infty$-groupoids indexed continuously by spaces, then you are essentially forced to notice the fundamental $\infty$-groupoid, and therefore also the higher homotopy groups.

By the way, the fundamental group $\pi_1$, which you seem to believe in already, appears in this story by restricting to 1-groupoids. It happens that the subcategory of 1-groupoids is reflective in $\infty$-groupoids, and the reflector applied to $\Pi X$ builds the usual "fundamental groupoid" $\Pi_1(X)$ whose isomorphisms are homotopy classes of paths in $X$. The universal property of this reflector means that $\Pi_1(X)$ carries all the information about families of objects of 1-groupoids parametrized continuously by $X$. In particular, applying this to the 1-groupoid of sets, we find that a "family of sets parametrized by $X$" is essentially a covering space of $X$, and so $\Pi_1(X)$ classifies covering spaces. And if $X$ is connected, then $\Pi_1(X)$ is essentially determined by $\pi_1(X)$.

To conclude, I must admit that the last four paragraphs began with a lie. There is no version of $\mathcal{T}$ that contains all of $\mathrm{Top}$ as a full subcategory and for which $\infty \mathrm{Gpd}$ is honestly reflective. There are versions of $\mathcal{T}$ (e.g. "continuous $\infty$-groupoids") that contain all "spaces built out of copies and open subsets of $\mathbb{R}^n$'s" and in which $\infty \mathrm{Gpd}$ is honestly reflective. There are also versions of $\mathcal{T}$ (e.g. "$(\infty,1)$-topoi") that contain all topological spaces, but in which the reflector ʃ is only defined on a (large) class of "nice" topological spaces. In the latter case, one can still define a "shape" ʃ $X$ of a non-nice space $X$ that "carries all the information about $X$-indexed families of objects of $\infty$-groupoids", but it is in general only a "pro-$\infty$-groupoid"; this includes for instance the information carried by the pro-discrete étale fundamental group.

Further reading (and shameless self-citations):

  • philosophical introduction to $\infty$-groupoids: section 2 of this paper
  • the above story told specifically about $(\infty,1)$-toposes: blog post
  • a similar version told for other versions of $\mathcal{T}$ (skip over the type theory): section 5 of this paper, and the introduction to this one
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    $\begingroup$ Nice answer,Mike. I avoided making explicit the distinction between spaces and $\infty$-groupoid on purpose, but thanks for doing so here. I would raise another question, and that is 'what is a space really?' I am posing that more or less 'philosophically' and certainly not just `mathematically', but it is a question that deserves to be discussed a bit more ... though here is not a suitable place! Certainly spaces are not the same as $\infty$-groupoids as you say. $\endgroup$
    – Tim Porter
    Oct 22, 2017 at 9:29
  • $\begingroup$ To be slightly contentious you start by saying: the modern realization that topological spaces are distinct from ∞-groupoids. As that realisation was inherent in both Borsuk's shape theory (1960s) and Grothendieck's letters to Breen (1975) I query 'modern', which in any case is a pretty flexible notion! The fact that most areas of algebraic topology historically avoided the spatial aspects of these other situations merely says that there were a lot of interesting and important areas in which they did not play a part, so they were sidelined 'for later study'. $\endgroup$
    – Tim Porter
    Oct 22, 2017 at 9:44
  • $\begingroup$ Thanks for this answer. I had been wondering about the relation between the infinity groupoid attached to the space and homotopy groups before asking the question. $\endgroup$
    – Asvin
    Oct 22, 2017 at 14:37
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    $\begingroup$ @TimPorter Sure. I might argue that although that realization may have existed in the literature since that time, it did not really percolate into practicing algebraic topology until much more recently. In particular, the clean description of the fundamental $\infty$-groupoid in topos-theoretic language didn't come about until the invention of $(\infty,1)$-topoi. $\endgroup$ Oct 22, 2017 at 21:37
  • $\begingroup$ Yes I agree with that, although the importance in geometric topology of the need for shape theoretic notions and in algebraic geometry for \'{e}tale homotopic' ones was in full view. $\endgroup$
    – Tim Porter
    Oct 23, 2017 at 5:56
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There are many things to say here. Here's one. Suppose you want to classify all spaces up to (weak) homotopy equivalence, or equivalently all CW complexes up to homotopy equivalence. The zeroth step is classifying the connected components of a space, so from now on let's assume that $X$ is connected. The first step is to understand the action of $\pi_1(X)$ on its universal cover; this is extremely complicated so let's ignore it entirely, and from now on let's assume that $X$ is simply connected.

$X$ can be understood by breaking it down into what is called its Postnikov tower; this is a sequence of maps

$$\cdots \to X_{\le 3} \to X_{\le 2} \to X_{\le 1}$$

where $X_{\le n}$ is the $n$-truncation of $X$, a space canonically built out of $X$ which has the same first $n$ homotopy groups but whose homotopy groups vanish above $n$. The relationship between the different stages of the Postnikov tower can be understood as follows. At each stage, there is a fiber sequence

$$B^{n+1} \pi_{n+1}(X) \to X_{\le n+1} \to X_{\le n}$$

where $B^{n+1} \pi_{n+1}(X) = K(\pi_{n+1}(X), n+1)$. For $n \ge 2$ this fiber sequence turns out to be classified by a map $X_{\le n} \to B^{n+2} \pi_{n+1}(X)$, or equivalently by a cohomology class

$$k_n \in H^{n+2}(X_{\le n}, \pi_{n+1}(X))$$

called the $n^{th}$ k-invariant of the Postnikov tower. These can be thought of as nonabelian analogues of classes in Ext groups classifying extensions.

In principle, then, one can hope to classify spaces by first classifying their homotopy groups and then classifying their k-invariants. Loosely speaking, what the Postnikov tower reveals is that

simply connected spaces are all "iterated extensions" of Eilenberg-MacLane spaces

and which Eilenberg-MacLane spaces appear is determined by the homotopy groups of $X$. This is generally hard to do for spaces but the same story can be applied to spectra and all sorts of other things, including chain complexes.

Example. Suppose we want to classify simply connected $3$-truncated spaces $X$; that is, spaces whose only nontrivial homotopy groups are $\pi_2$ and $\pi_3$. This is equivalent to various other classifications, namely

  • the classification of simply connected $3$-groupoids,
  • the classification of connected $2$-groups (after taking loop spaces),
  • the classification of grouplike braided monoidal groupoids (after taking loop spaces again).

A space $X$ with this property has a very simple Postnikov tower; it fits into a single fiber sequence

$$B^3 \pi_3(X) \to X \to B^2 \pi_2(X)$$

which is classified by a single k-invariant, namely a cohomology class in $H^4(B^2 \pi_2(X), \pi_3(X))$. This cohomology group can be very explicitly understood: it turns out to be precisely the group of $\pi_3$-valued quadratic forms on $\pi_2$. As an invariant attached directly to $X$, this quadratic form is a homotopy operation $\pi_2(X) \to \pi_3(X)$, represented by the Hopf map, which is a quadratic refinement of the Whitehead bracket.

For example, the $3$-truncation $\tau_{\le 3}(S^2)$ of the $2$-sphere is a simply connected $3$-truncated space with $\pi_2 = \pi_3 = \mathbb{Z}$. Hence it is classified by a quadratic form $\mathbb{Z} \to \mathbb{Z}$ which turns out to be the obvious guess $x \mapsto x^2$.

Above when I say "$3$-groupoid" I mean "weak $3$-groupoid." The difference between this and "strict $3$-groupoid" turns out to be that the k-invariant above always vanishes for a strict $3$-groupoid; the conclusion is that, for example, no strict $3$-groupoid models the homotopy type of the $3$-truncation of $S^2$.

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    $\begingroup$ So, one major difference between this story and the story of the fundamental group is that it is much more explicitly a story about homotopy types, whereas the fundamental group is also important at the level of spaces; e.g. the covering spaces of a manifold are still manifolds and so forth. I don't know any analogue of this sort of connection to geometric structures for the higher homotopy groups; typically the $n$-connected covers of a manifold, $n \ge 2$, can't be represented by manifolds, for example. $\endgroup$ Oct 21, 2017 at 18:51
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    $\begingroup$ Of course, if you don't know why to care about higher homotopy groups, then you probably don't know why to care about classifying things up to weak homotopy equivalence... $\endgroup$ Oct 21, 2017 at 22:30
  • $\begingroup$ @Qiaochu my thesis paved the road that would lead to the land where the 2-connected cover is a(n infinite-dimensional) Lie groupoid, equivalently a kind of differentiable stack. Higher analogues of this are doable with truncated internal Kan complexes. $\endgroup$
    – David Roberts
    Oct 22, 2017 at 8:34
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Many answers are going to seem at least a little circular, including this one. If you agree that CW complexes are an interesting class of spaces, encompassing any spaces you may want to consider, then the homotopy groups are exactly the invariants that tell spaces apart: A map $f\colon X\to Y$ from one CW complex to another is a homotopy equivalence if and only if it induces isomorphisms on all homotopy groups (allowing all possible basepoints if $X$ is not connected), i.e., is a weak equivalence. Of course, this also assumes that we are interested in spaces only up to homotopy equivalence, which is what algebraic topology generally assumes.

The reason I think of this as somewhat circular is that, underlying both the assumption that CW complexes are the most interesting spaces and the definition of higher homotopy groups is the belief that spheres are fundamental objects of study. CW complexes are built out of (cones on) spheres, and that pretty much directly leads to the fact that weak equivalences of CW complexes are homotopy equivalences.

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  • $\begingroup$ I agree, and this is probably going to be a problem for somebody who (as the questioner said) has a background in arithmetic geometry and often encounters topological spaces of a very different type. $\endgroup$ Oct 21, 2017 at 20:00
  • $\begingroup$ @TylerLawson if I am not mistaken, homotopy groups of finite topological spaces are isomorphic to the homotopy groups of the geometric realizations of the corresponding preorders. $\endgroup$ Oct 22, 2017 at 15:54
  • $\begingroup$ There is one added complication, Steve, and that given two spaces that you know the homotopy groups of, and yeah they are isomorphic, you still have to construct a suitable map realising the isomorphisms. This realisation problem was there from the start in J. H. C. Whitehead's fundamental Combinatorial Homotopy papers (late 1949s, early 1950s) and have been explored a lot by Baues in his books. The circularity that you point out is one reason why CW-complexes are where the rapid expansion of algebraic topology concentrated at least initially. $\endgroup$
    – Tim Porter
    Oct 23, 2017 at 6:06
  • $\begingroup$ Absolutely, regarding the complication. I didn't mention that, but stated the result carefully with it in mind. $\endgroup$ Oct 23, 2017 at 13:25
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    $\begingroup$ I think that since most of the "geometric" and "combinatorial" spaces that occur in mathematics - that is, manifolds, real and complex algebraic varieties, spaces built from combinatorial data like categories and posets - have the homotopy type of CW complexes, one can break the circularity so to speak by beginning with such objects as the motivating examples. Then they happen to fit into a nice class of objects - spaces of the homotopy type of a CW complex - for which higher homotopy groups are a very useful invariant. $\endgroup$
    – Dan Ramras
    Oct 23, 2017 at 21:56
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An axiomatic characterization of homotopy groups similar to that for (co)homology theories is given in Section 4 of Milnor's classic paper

Construction of Universal Bundles, I

John Milnor Annals of Mathematics Second Series, Vol. 63, No. 2 (Mar., 1956), pp. 272-284.

Apparently the idea that the existence of universal bundles implies such a characterization is due to Serre. It seems they are unique up to order of multiplication (the opposite groups also work).

In addition to that, stable homotopy groups have such an axiomatic characterization: they form the initial multiplicative homology theory.

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    $\begingroup$ Milnor did this. I'll try to track down the paper, but I think it was late 1950s or early 1960s. $\endgroup$
    – Jeff Strom
    Oct 21, 2017 at 20:17
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    $\begingroup$ Successive edits seem to have obscured the intended meaning of the last sentence. (What is being opposed to what?) $\endgroup$ Oct 22, 2017 at 9:44
  • $\begingroup$ @FrancoisZiegler tried to correct $\endgroup$ Oct 22, 2017 at 15:50
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    $\begingroup$ The stable homotopy groups are the homotopy groups of the sphere spectrum, which is an $E_\infty$-ring, and indeed the initial object in the $(\infty,1)$-category of ring spectra. $\endgroup$
    – David Roberts
    Oct 22, 2017 at 21:38
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Exact sequences occur in many parts of mathematics. The shorter ones can arise in many different ways. However, if you have an exact sequence of length six or more, it can usually be interpreted naturally using homological algebra, as the long exact sequence associated to a short exact sequence of chain complexes. There are some sequences that cannot be interpreted that way, but essentially all of those can be understood as the long exact sequence of homotopy groups associated to a suitable fibration. Typically, there is no other useful way to understand them. Moreover, the Dold-Kan theorem allows us to convert any short exact sequence of chain complexes to a fibration of spaces. Thus, the theory of fibrations and homotopy groups is essentially the simplest context in which we see all naturally occurring long exact sequences.

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In my opinion, the motivation is that the set of (path) connected components $\pi_0(X)$ of a space and the loop space $\Omega(X,x_0)$ of a pointed space are two important properties that respect homotopy types.

Since they are both important, we might expect their iterated combinations to be important as well, such as $\pi_0(\Omega^n(X, x_0))$. Of course, $\pi_0(\Omega^n(X, x_0))$ is simply $\pi_n(X, x_0)$.


In my opinion, this point of view also leads to an explanation of why weak homotopy equivalence is the "right" notion of equivalence, which puts further emphasis on the higher homotopy groups.

Suppose we wanted to describe a condition that $X \simeq *$. That is, $X$ has a unique point, but uniqueness should only be up to unique homotopy (where uniqueness is only up to unique homotopy, and so forth).

We can split the requirement that $X \simeq *$ into two parts:

  • Any two points are equivalent
  • For any two points, $\mathrm{Path}(x,y) \simeq *$

Since $\mathrm{Path}(x,y)$ is homotopy equivalent to $\Omega(X,x)$, this reduces to

  • $X$ is path connected
  • $\Omega X \simeq *$. (I've omitted the base point since its choice doesn't matter)

This recursive definition leads to a necessary condition that $\pi_0(X) \cong 0$ and $\pi_n(X, x_0) \cong 0$ for all $n>0$.

If we want failure of equivalence to be detectable by homotopies or by homotopies between homotopies or by homotopies between homotopies or so forth, then we need this to be a sufficient condition as well.

Then if we say that $f : X \to Y$ is an equivalence iff its homotopy fibers are equivalent to a point, and the long exact sequence reduces this condition to $f$ inducing an isomorphism $\pi_0(X) \to \pi_0(Y)$ and $\pi_n(X,x_0) \to \pi_n(Y,f(x_0))$ for all $n>0$ and all basepoints.

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A short answer might be that your point 1 relates rather to the fundamental group(oid) being enough to determine the 1-type of the space and this is more or less the Galois theoretic viewpoint. Even for the 2-type of a space there is not a simple equivalent formulation to the covering space, but there is a model, a crossed module, which encodes the 2-type (so that includes both $\pi_1$ and $\pi_2$ and the relevant Postnikov invariant). Defining the standard fundamental crossed module of a CW-complex, $X$, is easy. It just uses the (relative) homotopy groups of the skeleta of $X$. So using a slight variant (the relative form of the usual one) of the second homotopy group, you get a good analogue of the fundamental group in the next dimension ... and you do get a version of the covering space classification, but it is more complicated.

The discussion of the 3-type etc. is again more complicated and for an idea of where that leads, look at the first part of Grothendieck's Pursuing Stacks manuscript (The letter to Quillen) and if you can, also have a look at the earlier (1975) letters to Larry Breen. Grothendieck saw that the covering space analogue gave an excellent way of extending from 2 or 3 dimensions to higher dimensions and, quite naturally, the fibration long exact sequence and the other features mentioned in the excellent replies from others, comes into the picture. This links up fibrations with fibres an (n-1)-type and representations of algebraic \categorical models of the n-type of $X$.

Now this does not directly use maps from a n-sphere into $X$, but look at Grothendieck's sketch of his construction of the fundamental $n$-groupoid of $X$, it uses a neat idea of using n-globes to define something like a relative homotopy group in each dimension, in other words, the basic building blocks are given by maps of n-spheres / n-globes into $X$ which again makes the link with the other replies. (For a filled out version of Grothenck's sketch see work by George Maltsiniotis.)

(I should point out that the notion of homotopy n-type is key to all this, and that in its easiest formulation involves the homotopy groups. Grothendieck's intuition is that the analogue of covering spaces / 0-stacks corresponds to the homotopy $n$-type, not just the single $\pi_n$ of the space. A negative feature of the usual definition of $\pi_n$ is that for `spaces' that occur in algebraic and arithmetic geometry there are VERY FEW maps from an $n$-sphere to the space, so points about loop spaces, although very important, cannot be directly applied to those spaces. In fact the link with Galois theory and SGA1-style fundamental groups has the advantage of not requiring the space to be a space at all, nor to have a CW-complex structure if it is a space, it could be a topos, or more generally an $n$-topos, in the sense of Lurie, but that probably leads too far from the initial question, and I did say this would be a 'short answer'!)

The full picture is not yet in place for how $n$-stacks are best to be used, even in low dimensions, so higher Galois theory for instance exists but its interpretation and application is much less advanced (quite naturally) and there is a lot more work to be done there.

I hope this helps.

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When the homotopy groups were announced at the Zurich ICM by E. Cech in 1932 they were regarded as quite wrong, since they were abelian: topologists of the time were hoping for nonabelian higher dimensional analogues of the fundamental group, in order to model the increasing complications of higher dimensions. For some discussion of this, see this paper Modelling and Computing Homotopy Types:I, to appear in Indagationes in 2017.

For a discussion of using kinds of $n$-fold groupoids to model pointed homotopy $n$-types, and do some calculations using a higher order van Kampen Theorem, see this expository paper.

One theme of these papers is dealing with "structured spaces"; see also "Esquisse d'un Programme", Section 5. Picking a point in a space is a rather minimalist kind of structure!

Sorry I am short of time to be more explicit!

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I feel a very simple argument in favor of higher homotopy groups is the invariance of domain: it had been an important problem to prove that $\mathbb{R}^n$ and $\mathbb{R}^p$ are not homeomorphic when $n\neq p$, and the homotooy groups of the complement of a point in each of these spaces seems now the most obvious way to prove this.

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    $\begingroup$ You can easily prove invariance of domain using homology groups instead... $\endgroup$ Oct 24, 2020 at 18:37
  • $\begingroup$ @DenisNardin: sure, but is it obviously simpler? $\endgroup$ Oct 24, 2020 at 20:06
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    $\begingroup$ Simpler is in the eye of the beholder, but I would argue that computing $H_n(S^n)$ is much easier than $\pi_nS^n$ (that requires some proof of Freudenthal's suspension theorem -- say with Blakers-Massey). In general I find it very hard to argue that Blakers-Massey is a "simpler" result than the Mayer-Vietoris les. $\endgroup$ Oct 24, 2020 at 20:41

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