[EDIT: This question as written contains a glaring error. It's sort of "not a real question"? But I am resisting the temptation to delete it, because I am going to make it into a real question.]

I tend to pay a lot of attention to when basepoints matter and when they don't. Currently something is bothering me:

Consider the "homotopy category" of unbased $1$-connected (i.e. simply-connected) spaces: that is, start with spaces that have one path component and trivial fundamental group, with unbased maps as the morphisms, and invert the weak homotopy equivalences. Or, if you prefer, start with $CW$ spaces of that kind and pass to homotopy classes. I suppose that this is NOT actually (equivalent to) the homotopy category of any model category, which is why I have used quotation marks above.

[EDIT: As Fernando Muro points out in his answer, the previous sentence is obviously wrong. The model structure described in the following paragraph, with its artificial use of a basepoint, does the job!]

If I do the same thing with basepoints, there's no problem. For example, I can take the usual category of based spaces but declare the weak equivalences to be the maps inducing isomorphisms of $\pi_j$ for all $j\ge 2$ at the basepoint. Then every object is equivalent to a $1$-connected space (in fact, if you use the usual fibrations, then all cofibrant objects are $1$-connected), and the homotopy category is as intended. The same kind of thing works for $k$-connected based spaces for other values of $k$.

When $k=0$ it is easy to see that the "homotopy category" of $k$-connected (i.e. path-connected) unbased spaces is not a homotopy category of a model category. In fact, it lacks coproducts. (Exercise: there is no universal example, up to homotopy, of a path-connected space equipped with two maps from the circle.) This is enough, because the coproduct (colimit of discrete diagram) of cofibrant objects in a model category is always a homotopy colimit.

When $k=-1$ (so that $k$-connected spaces means nonempty spaces) it is even easier to see: The homotopy category of nonempty spaces has no initial object.

[EDIT: I will leave this as it stands even though the answer is, come to think of it, obviously "no".]

QUESTION 1: Is it true that the homotopy category of (unbased) $1$-connected spaces cannot be the homotopy category of a model category?

This is not so obvious. You can't just say something like "a homotopy category must have all limits and colimits" because homotopy limits and colimits are not in fact limits and colimits in a homotopy category in general -- the homotopy category of spaces does not have pullbacks.

(I presume that most of what I said above more or less applies to other popular ways of axiomatizing homotopy theory, such as $\infty$-categories.)

Apart from this technical question, I have a vague philosophical question:

QUESTION 2: If the answer to Question 1 is "yes", what is the right response?

I am thinking of rational homotopy theory, for example. I recall that when Quillen worked out an indirect equivalence between on the one hand $1$-connected spaces and rational equivalences and on the other hand $0$-connected commutative differential graded coalgebras over $\mathbb Q$ he used based spaces and coaugmented coalgebras. It's clear why. But Sullivan's equivalence between rational homotopy theory of simply connected spaces and commutative dgas works without basepoints, right? He doesn't do it in a model category framework and (this is really beside the point) he needs finite type conditions. It seems unnatural to insist on based spaces just for the sake of (model category or other) axioms. On the other hand they are such nice axioms ...

  • $\begingroup$ I don't see why question 1 is "obvious", at least phrased in this way, which only puts a condition on the homotopy category of a model category. This is related to the question: does there exist model category $M$, such that $hM$ and $h\mathrm{Top}$ are equivalent as categories, but $M$ is not Quillen equivalent to $\mathrm{Top}$? If you don't impose some additional structure (like cofiber sequences) for the equivalence $hM\approx h\mathrm{Top}$ to preserve, this is hard to answer (and maybe it's false, I don't know). $\endgroup$ Mar 7 '12 at 19:10
  • $\begingroup$ Charles: I mean that the answer is obviously that there is a model category whose homotopy category is equivalent to this. But on the other hand, so what? $\endgroup$ Mar 8 '12 at 0:49
  • $\begingroup$ @Tom: I don't quite understand what you are trying to ask. Nevertheless, I will leave the following remark, just in case. The quasi-category of 1-connected spaces (full sub-quasi-category of the quasi-category of spaces) does not have any binary coproducts. To see this, note that for any 1-connected space $X$ the space of maps $\text{Map}(X,K(G,2))\simeq K(G,2)\times H^2(X,G)$ has $\pi_2=G$ (for $G$ an abelian group). However, if a binary coproduct $X\coprod Y$ existed in 1-connected spaces, the space of maps $\text{Map}(X\coprod Y,K(G,2))=\text{Map}(X,K(G,2))\times\text{Map(Y,K(G,2))}=\cdots$ $\endgroup$ Mar 8 '12 at 11:10
  • $\begingroup$ $\cdots =K(G,2)^2\times H^2(X,G)\times H^2(Y,G)$ would have $\pi_2=G\times G$ which would contradict the previous statement in the case $G=\Bbb{Z}$. $\endgroup$ Mar 8 '12 at 11:24

I may be missing something, but let me throw some ideas:

When you say 'homotopy category' in Q1, I'll take it as the quotient of the category of CW-complexes with only 1-vertex and no 1-cells by the unbased homotopy relation. By cellular approximation, homotopy extension, etc. unbased homotopy classes of maps between such spaces coincide with based homotopy classes of cellular maps. Therefore it would be enough to have a model structure on the category of 1-connected pointed spaces with $\pi_*$-isomorphisms as weak equivalences. This is achieved in:

MR1353370 (97a:55009) Extremiana Aldana, J. Ignacio; Hernández Paricio, L. Javier; Rivas Rodríguez, M. Teresa A closed model category for (n−1)-connected spaces. Proc. Amer. Math. Soc. 124 (1996), no. 11, 3545–3553. (Reviewer: Timothy Porter)

Of course, they are much inspired by Quillen's work on the simplicial group counterpart in his "Rational homotopy theory" paper. See also:

MR1387446 (97a:55022) Garzón, Antonio R.; Miranda, Jesús G. Closed model structures for algebraic models of r-connected spaces. C. R. Math. Rep. Acad. Sci. Canada 18 (1996), no. 1, 27–32

  • $\begingroup$ You are right, of course. I know this model very well; I just foolishly did not notice that when $n>0$ this model for $n$-connected based spaces is also a model for $n$-connected unbased spaces. I will have to edit the question! $\endgroup$ Mar 7 '12 at 11:40
  • $\begingroup$ I mean that it is a model for the homotopy category of $n$-connected based spaces, which is what I asked. But it's not a model for the homotopy theory of such spaces, which is what I ought to have asked. $\endgroup$ Mar 8 '12 at 2:02

I'll throw in a point (!) I have been making since 1967 that certainly for the 1-dimensional case it is convenient to use a set of base points chosen according to the geometry. Thus one finds that in van Kampen type situations one is calculating homotopy 1-types: groupoids are useful because they have structure in dimensions 0 and 1.

Analogously, in higher dimensions one needs algebraic objects with structure in a range of dimensions, especially as low dimensional identifications have high dimensional influence on homotopy types. It turns out that this can be accomplished to a useful extent if one works with filtered spaces, or $n$-cubes of spaces. With these structures, one can calculate precisely some homotopy $n$-types in gluing situations.

The implications of having many base points are unclear in, for example, the theory of loop spaces, and their iterations, or for model categories. But it is clearly sensible to have a set of base points in discussing group actions, where the action group $G$ acts also on the fundamental groupoid $\pi_1(X,A)$ if and only if $A$ is union of orbits. Thus the natural theorem on orbit spaces involves orbit groupoids.

  • 3
    $\begingroup$ I always like your answers, they are interesting, but this one does not really address the question - except the unifying theme of base points ... $\endgroup$ Mar 7 '12 at 15:55
  • $\begingroup$ In my experience the notion of "base point" reflects some underlying and wider issues in algebraic topology, namely algebraic modelling and calculation. I originally thought the fundamental groupoid was great as it enabled getting rid of base points. I gradually realised that to calculate e.g. the fundamental group of the circle one needed a set of base points. In higher dimensions one needed some more structure. Compare Grothendieck "Esquisse .." Section 5. Choosing a base point enabled some group structures. If these turn out inadequate, one needs to look at other choices- so... $\endgroup$ Mar 8 '12 at 10:36

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.