Why the homotopy groups are groups, I

I’ve been reading a lot of algebraic topology as of late, but I still don’t think I understand most of it properly. I will try to blog about some of the ideas as a means to understand the ideas better.

The homotopy groups are essentially homotopy classes of maps from the sphere into a space ${X}$ . For instance, the first homotopy group ${\pi_1}$ (or fundamental group) is the group of homotopy classes of loops at a fixed base-point. But why should they be a group? There is a categorical reason for this, and while it’s not immensely deep, I’d like to explain it.

1. Definition

Fix a topological space ${X}$ with a basepoint ${x_0 \in X}$ . Let ${n \geq 1}$ . We consider the set ${[I^n : X]}$ of homotopy classes of maps ${I^n \rightarrow X}$ from the unit ${n}$ -cube into ${X}$ . This, however, is not a very interesting set, because ${I^n}$ is contractible; there is thus only one element. Instead, we consider the set ${[I^n, \partial I^n: X, x_0]}$ of homotopy classes of maps ${I^n \rightarrow X}$ that send the boundary ${\partial I^n}$ into ${x_0}$ . When I say “homotopy classes,” I mean that homotopies are required to send the boundary ${\partial I^n}$ into ${x_0}$ .

To make this explicit:

Definition 1 The ${n}$ -th homotopy group ${\pi_n(X, x_0)}$ of a space-with-basepoint ${(X, x_0)}$ is the set of all maps ${I^n \rightarrow X}$ sending ${\partial I^n}$ to ${x_0}$ under the equivalence relation that ${f, g: I^n \rightarrow X}$ are considered equivalent if there exists a homotopy $\displaystyle H: I^n \times I \rightarrow X$

between ${f,g}$ relative to ${\partial I^n}$ . This is functorial in the obvious way: a map ${(X, x_0) \rightarrow (Y, y_0)}$ (i.e. a map sending ${x_0 \rightarrow y_0}$ ) induces a map on the homotopy groups.

So this is a set. But we have not explained why this is supposed to be a homotopy group. The group structure can be visualized intuitively in the case ${n=1}$ : given ${f, g: I^1 \rightarrow X}$ that each send the boundaries into ${x_0}$ , we can catenate ${f,g}$ to get a path

$\displaystyle h(t) = \begin{cases} f(2t) & \mathrm{if} \ t \leq \frac{1}{2} \\ g(2t-1) & \mathrm{if} \ t \geq \frac{1}{2}. \end{cases}$

This multiplication is not associative on the loops, but it is associative up to homotopy. And that’s all you need to define the group law on the set of homotopy classes when ${n=1}$ . A modification of this works in higher dimensions.

The essential point here was that paths can be pieced together back to back. This is an essential point of the space ${I^1}$ . For instance, it would be less obvious how to make the set of homotopy classes from a pointed torus into a space a group in any natural way.

The answer to the earlier question is that ${I^1/\partial I^n = S^1}$ (and the higher ${I^n/\partial I^n = S^n}$ ) is “almost” a co-group object in the category of pointed topological spaces. I say almost, because the comultiplication is (co)associative only up to homotopy. In particular, the fact that all these homotopy groups are groups is a special case of Yoneda categorical nonsense.

2. The homotopy category

Since we work with maps up to homotopy to define the groups, it makes sense to consider a corresponding category. In this category, the functors ${\pi_1, \pi_2, \dots}$ are representable by the objects ${I^n/\partial I^n}$ . Thus the fact that they can be viewed naturally as groups is a co-group statement about the ${I^n/\partial I^n}$ .

Definition 2 We consider the category ${\mathbf{PT}}$ , called the homotopy category of pointed topological spaces, defined as follows. The objects of ${\mathbf{PT}}$ are pointed spaces ${(X, x_0)}$ . So ${X}$ is a topological space and ${x_0 \in X}$ .

The morphisms between ${(X, x_0), (Y, y_0)}$ are the set of maps ${f: X \rightarrow Y}$ with ${f(x_0) = y_0}$ with the equivalence relation that ${f,g}$ are equivalent if there is a homotopy ${H: X \times I \rightarrow Y}$ between ${f,g}$ such that ${H(x_0 \times I) = y_0}$ .

The last condition says that ${H}$ is a homotopy relative to ${x_0}$ . Strictly speaking, one should check something first. Namely, that if ${f,g: X \rightarrow Y}$ are equivalent and ${h: Y \rightarrow Z}$ is a pointed map, then ${h \circ f, h \circ g}$ are equivalent; this is necessary to see that the category is well-defined. But this is straightforward.

So what is the set ${\pi_n(X, x_0)}$ in this language? Recall that a map ${I^n \rightarrow X}$ sending ${\partial I^n \rightarrow x_0}$ is the same thing as a map out of the quotient space ${I^n/\partial I^n \rightarrow X}$ sending the “basepoint” ${\partial I^n \in I^n/\partial I^n}$ . So if we regard ${(I^n/\partial I^n, \partial I^n)}$ as a pointed space, then ${\pi_n(X, x_0)}$ is just

$\displaystyle \hom_{\mathbf{PT}}( (I^n/\partial I^n, \partial I^n), (X, x_0)).$

It is clear that ${\pi_n(X, x_0)}$ is a covariant functor from ${\mathbf{PT} \rightarrow \mathbf{Sets}}$ . It is also clear that it is co-representable by the object ${(I^n/\partial I^n, \partial I^n) \in \mathbf{PT}}$ . But in fact, as asserted above, each ${\pi_n(X, x_0)}$ is a group.

Proposition 3 Let ${\mathfrak{C}}$ be any category with finite coproducts and a final object. Suppose ${X \in \mathfrak{C}}$ and the sets ${F_X(Y) = \hom_{\mathfrak{C}}(X, Y)}$ are endowed with a group structure which is natural. In other words, ${\hom_{\mathfrak{C}}(X,Y)}$ can be viewed as a functor to the category of groups. Then ${X}$ is a co-group object.

Let us recall what it means for ${X}$ to be a co-group object. In other words, there is a comultiplication map

$\displaystyle X \rightarrow X \sqcup X$

and a counit map

$\displaystyle \ast \rightarrow X$

for ${\ast \in \mathfrak{C}}$ the final object. These are required to satisfy arrow-theoretic reversals of the usual relations imposed on a group. For instance, the map

$\displaystyle X \rightarrow X \sqcup X \rightarrow X \sqcup \ast \rightarrow X$

composing the counit and the comultiplication, as to be the identity.

Now ${\mathbf{PT}}$ has a final object: it’s just ${\ast = (x_0, x_0)}$ , a one-point space. Moreover, it has finite coproducts. The coproduct of ${(X, x_0), (Y, y_0)}$ is the space ${X \sqcup Y/\left\{x_0 \sim y_0\right\}}$ . This is not completely trivial, actually. On the one hand, it is clear that to give a pointed map of topological spaces

$\displaystyle X \sqcup Y/\left\{x_0 \sim y_0\right\} \rightarrow (Z, z_0)$

is the same as giving a pair of pointed maps ${(X, x_0), (Y, y_0) \rightarrow (Z, z_0)}$ . We need to check, however, that a pair of homotopies

$\displaystyle H': (X, x_0) \times I \rightarrow (Z, z_0), \quad H'': (Y, y_0) \times I \rightarrow (Z, z_0)$

induce

$\displaystyle H: (X \sqcup Y/\left\{x_0 \sim y_0\right\}, \left\{x_0, y_0\right\} )\times I \rightarrow (Z, z_0).$

The reason is that ${H, H'}$ piece together to give a map of the quotient space of ${X \times I}$ and ${Y \times I}$ with ${(x_0, t) \sim (y_0, t)}$ for each ${t \in I}$ . By the following theorem in general topology, this quotient space is the same as the product ${(X \sqcup Y)/\left\{x_0 \sim y_0 \right\} \times I}$ . So the homotopies indeed do patch appropriately.

Lemma 4 Let ${X \rightarrow Y}$ be a quotient map. If ${Z}$ is a locally compact Hausdorff space, then ${X \times Z \rightarrow Y \times Z}$ is a quotient map.

We won’t prove this nontrivial but not terribly difficult result from general topology.

So, let’s put this together. ${\mathbf{PT}}$ has finite coproducts and a final object. So when a (co)representable functor ${\hom_{\mathbf{PT}}(Z, -)}$ has its image in the category of groups, then ${Z}$ is a cogroup in ${\mathbf{PT}}$ . Unravelling the definitions shows that this means that there are continuous maps of pointed topological spaces

$\displaystyle Z \rightarrow Z \sqcup Z$

and

$\displaystyle Z \rightarrow \ast$

such that coassociativity holds up to pointed homotopy.

Definition 5 A H cogroup is a pointed topological space ${(Z, z_0)}$ with comultiplication and counit maps such that the cogroup diagrams commute up to pointed homotopy. In other words, a cogroup object in ${\mathbf{PT}}$ .

So we see that necessarily ${I^n/\partial I^n = S^n}$ is an ${H}$ cogroup, which is what lets one catenate a pair of maps ${(I^n, \partial I^n) \rightarrow (Z,z_0)}$ . This cogroup structure in the case ${n=1}$ , for instance, sends ${t \in I/\left\{0,1\right\}}$ to ${2t}$ in the first copy of ${I/\left\{0,1\right\}}$ if ${t \leq \frac{1}{2}}$ and to ${2t -1 }$ in the second copy of ${I/\left\{0,1\right\}}$ for ${t \geq \frac{1}{2}}$ . Since the boundaries of the two are identified, this is a well-defined comulitiplication map, and one can check that it is homotopy co-associatve. The counit collapses everything to a point.

And that is one basic reason for why the homotopy groups should be groups.

This still doesn’t explain why we would expect ${S^n}$ to be an ${H}$ cogroup. Next, I will discuss a pair of adjoint functors on ${\mathbf{PT}}$ that helps answer this related question, as well as the further question of why the higher homotopy groups are commutative.

Why the homotopy groups are groups II « Climbing Mount Bourbaki Says:

September 26, 2010 at 11:38 pm

[…] 2010 tags: adjoint functors, homotopy theory, loop space, reduced suspension by Akhil Mathew Last time, we were discussing the category whose objects are pointed topological spaces and whose morphisms […]

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