7
$\begingroup$

I believe that $S^1\vee S^1$ is the Eilenberg-Mac Lane space $K(\mathbb{Z}\ast\mathbb{Z},1)$. One can prove this by constructing its universal cover and observing that it is contractible.

My question is this:

Is there a simple homotopy-theoretic proof of this result that does not use connected covers?

I have never seen one appear in any text. Repeated attemps by myself to use the usual tools (Hilton-Milnor Theorem, examine the fibre of the inclusion $S^1\vee S^1\hookrightarrow S^1\times S^1$, etc...) to prove this elementary result have met with connectivity issues or similar problems.

$\endgroup$
  • 2
    $\begingroup$ It's not clear to me what kind of answer you're after. Presumably such a proof would take as input that a circle is a K(Z,1). But how do you prove that without mentioning covering spaces? $\endgroup$ – Dan Petersen Jul 11 '16 at 15:12
  • 13
    $\begingroup$ I think that a homotopy purist would first want to formulate the question as follows: If $X$ is a K($G$, 1) and $Y$ is a $K($H$, 1)$, is $X\vee Y$ a K($G*H$, 1)? This is true, but I don't know how to prove it without using universal covering spaces. Searching around a little on Google, I didn't find any discussion of this fact, apart from one paper stating it without proof. But maybe I haven't looked hard enough. $\endgroup$ – Dan Ramras Jul 11 '16 at 15:57
  • 2
    $\begingroup$ Maybe a homotopy purist would say that there's nothing very homotopically pure about the fundamental group (e.g. it's nonabelian), and hence we shouldn't expect this to follow from very general homotopical considerations. Or maybe one will come along with a nice proof! $\endgroup$ – Dan Ramras Jul 11 '16 at 16:41
  • 2
    $\begingroup$ I also wish I knew how to prove this without using covering spaces. The most general fact I know about things being unexpectedly 1-types is that if you have a functor $F$ from the free category on a directed graph $G$ to the category of groups and injective group homomorphisms, then the homotopy colimit of the classifying spaces of those group, $\mathrm{hocolim}_{x \in G} BF(x)$, is a 1-type. This is Theorem 1B.11 in Hatcher's Topology. (The proof builds the universal cover of the colimit.) $\endgroup$ – Omar Antolín-Camarena Jul 11 '16 at 19:29
  • 8
    $\begingroup$ I mildly object to the insinuation that a pure homotopy theorist wouldn't like the universal cover approach. I mean here you don't need the universal cover up to homeomorphism, only up to homotopy equivalence. That's certainly pure homotopy theory: if $X$ is connected, the universal cover of $X$ has the homotopy type of the homotopy fibre of the canonical map $X \to \tau_{\le 1} X$. Jesse's Ans 1 shows how to prove that this fibre is contractible without using covering space theory. $\endgroup$ – Omar Antolín-Camarena Jul 12 '16 at 18:05
6
$\begingroup$

I think the following can be turned into a proof, but I haven't checked the details.

By a result of Milnor, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $F(S^0 \vee S^0)$, the free group functor on the pointed set $S^0 \vee S^0$. By a version of the Hilton-Milnor theorem, the latter coincides up to homotopy with $F(S^0) \ast F(S^0)$, the latter being the free product with amalgamation of simplicial free groups. But $F(S^0)$ coincides with $\Bbb Z$ up to homotopy. Hence, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $\Bbb Z \ast \Bbb Z$ (as a space). In particular, $\Omega (S^1 \vee S^1)$ is homotopically discrete, which gives the result you are after (in conjunction with the van Kampen theorem).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Have you got a reference for Milnor's result? $\endgroup$ – Tyrone Jul 12 '16 at 14:20
  • 1
    $\begingroup$ It appears in Adams, John Frank Algebraic topology—a student's guide. London Mathematical Society Lecture Note Series, No. 4. Cambridge University Press, London-New York, 1972 $\endgroup$ – John Klein Jul 13 '16 at 3:26
3
$\begingroup$

$\renewcommand{\deloop}{\mathbf{B}}\renewcommand{\sph}{\mathbb{S}} $

Ans 1: Calculate the homotopy fiber of the map $\sph^1\vee \sph^1\to \deloop F_2$ classifying the generators. This (by distributivity/Mather's Cube Theorems) is the pushout of the two maps $\langle x,y\rangle \to \langle x,y\rangle/\langle x\rangle$ and $\langle x,y\rangle \to \langle x,y\rangle/\langle y\rangle$; that this pushout is contractible boils down to the combinatorial exercise that a word in $\{ x,y \}$ knows what its letters are. (If you like, this is the homotopy reason for the connected cover story; it may also be what Qiaochu is getting at, but I'm not quite sure I can tell for sure)

Ans 2: The loopspace functor $\Omega$ from connected pointed spaces $\mathrm{Top}_{0,*}$ to (tame-enough) grouplike $A_\infty$-spaces has a homotopy inverse $\mathbf{B}$.

In particular, $\deloop$ sends colimits to colimits, and $\deloop\mathbb{Z}\simeq\sph^1$.


Ans 2 may sound fancier than Ans 1, but they're actually built from the same tools.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ For Ans 2, how do you show that the coproduct of $\mathbb{Z}$ and $\mathbb{Z}$ is $\mathbb{Z} * \mathbb{Z}$ in grouplike $A_\infty$-spaces? I know it's the coproduct in groups (or even monoids), can that statement be reduced to this one somehow? $\endgroup$ – Omar Antolín-Camarena Jul 11 '16 at 21:09
  • 1
    $\begingroup$ Right, Omar's question is the reason this isn't a formal exercise (at least as far as I can see). Analogously to my answer, the inclusion of groups into grouplike $E_1$ spaces has a left adjoint given by $\pi_0$, so it preserves homotopy limits for formal reasons, but the question at hand is why it preserves a certain homotopy colimit. $\endgroup$ – Qiaochu Yuan Jul 11 '16 at 21:38
  • 1
    $\begingroup$ The spaces in the $A_\infty$-operad are contractible? ~~~~ Or, what happens if I say: let $G$ and $H$ be simplicial groups, and define $(G * H)_k = (G_k) * (H_k)$ with faces/degeneracies defined extending $G_k \to G_{k\pm 1} \to (G * H)_{k\pm 1} $ ... (which are therefore simplicial groups again)? ~~~~ Good questions, of course. Will get back to you. $\endgroup$ – Jesse C. McKeown Jul 11 '16 at 21:48
  • 1
    $\begingroup$ $A_\infty$ spaces can be strictified. So when both exist in the right category, the $A_\infty$-colimit is a competitor, in groups, to the group colimit, and the group colimit is a competitor, in $A_\infty$-spaces, to the $A_\infty$-colimit. Composing comparison maps on both sides et.c. $\endgroup$ – Jesse C. McKeown Jul 11 '16 at 21:52
  • $\begingroup$ I like your Ans 1! Although I have to say it felt very much like building the universal cover and observing it's a tree. What I mean is that the only way I saw to show the contractibility of the homotopy pushout of discrete spaces $F_2/\langle x\rangle \leftarrow F_2 \rightarrow F_2/\langle y \rangle$ is to prove that the usual double mapping cylinder construction of the homotopy pushout is a tree. $\endgroup$ – Omar Antolín-Camarena Jul 11 '16 at 22:18
3
$\begingroup$

This is not really an answer. The inclusion of homotopy $1$-types into homotopy types has a left adjoint $\tau_{\le 1}$ given by truncation, and hence preserves homotopy limits (e.g. products). The question at hand is whether there is a homotopy-theoretic way to see that it also preserves certain homotopy colimits, namely a certain shape of homotopy pushout.

One might ask more generally what happens when $1$ is replaced by $n$, and here I believe it's not true that the inclusion of homotopy $n$-types preserves homotopy pushouts; e.g. I don't believe it's true that the wedge sum of two homotopy $n$-types is another homotopy $n$-type. (The first counterexample that comes to mind only works rationally: rationally $S^2$ is a $2$-type, but $S^2 \vee S^2$ has rational homotopy in arbitrarily high degrees.)

So there's something special about the inclusion of homotopy $1$-types. At this point you might be happy to learn that it's possible to develop covering space theory for groupoids without any reference to topological spaces. I think this is already enough to prove the desired result, by a construction, at the level of groupoids, of the universal cover of $BG \vee BH$. In fact it ought to be possible to completely describe the covering theory of $BG \vee BH$ at the level of groupoids, leading to the Kurosh subgroup theorem. For some indication of how this might be done see this blog post, which has some pictures of the special case $G = C_2, H = C_3$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The most general homotopy colimit I know that $\tau_{\le 1}$ preserves is given by Theorem 1B.11 in Hatcher's book. It gives you that $\mathrm{hocolim} BF$ is a 1-type if $F : \mathcal{C} \to \mathrm{Groups}$ is a diagram of groups and injective group homomorphisms and $\mathcal{C}$ is the free category on a directed graph. $\endgroup$ – Omar Antolín-Camarena Jul 11 '16 at 19:36
  • 1
    $\begingroup$ @Omar: I know an example not covered by that statement; the inclusion of $1$-types (not $\tau_{\le 1}$, which is its left adjoint) also preserves homotopy quotients by (discrete) group actions. $\endgroup$ – Qiaochu Yuan Jul 11 '16 at 19:37
  • 3
    $\begingroup$ In fact we always have $\pi_{2n-1}(K(A,n) \vee K(B,n)) = A \otimes B$ for $n > 1$, and so this is often not an $n$-type. Something genuinely special is happening in the case $n=1$. $\endgroup$ – Tyler Lawson Jul 11 '16 at 20:25
  • $\begingroup$ A groupoid proof of the Kurosh subgroup theorem is in Philip Higgins' 1971 book "Categories and Groupoids" available at tac.mta.ca/tac/reprints/articles/7/tr7abs.html . Philip told me he thought of the use of groupoids when reading Hilton and Wylie's account of covering spaces. $\endgroup$ – Ronnie Brown Jul 12 '16 at 10:05

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.