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I apologize in advance if this the answer to this question is standard or well-known. I am not in any way an algebraic topologist.

$\newcommand{\s}{\mathscr}$Let $\s T$ be the category of topological spaces, $\s T_*$ the category of pointed topological spaces. We can construct the homotopy category $\s H$ and the "pointed homotopy category" $\s H_{*}$ by letting $\hom_{\s H}(X,Y)$ be $\hom_{\s T}(X,Y)$ modulo homotopy and $\hom_{\s H_*}(X,Y)$ be $\hom_{\s T_*}(X,Y)$ modulo homotopies preserving basepoints. There are obvious functors $F:\s T\to \s H$, $F_{*}:\s T_{*}\to \s H_{*}$.

My questions are pretty basic: do $F,F_*$ preserve products? arbitrary limits? only arbitrary finite limits? coproducts? arbitrary colimits? It not, how does one compute colimits in $\s H$ and $\s H_{*}$?

If this can be found in a standard reference, just write an answer saying "this is a stupid question - this is all worked out in...," and I'll upvote and accept.

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Products and coproducts, yes. Most other limits and colimits, no. I recommend it as exercise to see why and why not. A bit more subtle: many diagrams in the homotopy categories do not even have limits or colimits. I don't have a reference. –  Tom Goodwillie Feb 1 '13 at 19:15
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From my spectator's vantage point, the most intriguing property of the homotopy category is its nonconcretizability. This is a theorem of Freyd, discussed in tac.mta.ca/tac/reprints/articles/6/tr6abs.html. He begins with a criterion involving kernels (normal subobjects), but immediately remarks (page 3): "There are very few kernels, indeed there are very few monomorphisms in H." This remark is where my own wild-goose chase began. The remainder of the article develops a more nuanced criterion involving sufficiently prevalent "generalized normal subobjects". –  Adam Epstein Feb 2 '13 at 11:41
    
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4 Answers

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For products and coproducts, this is covered on page 67-68 of "Introduction to Homotopy Theory" by Martin Arkowitz

To see that the homotopy category is not complete or cocomplete (i.e. that there are diagrams which don't have limits or colimits), check out "Modern Classical Homotopy Theory" by Jeffrey Strom, page 435. He explicitly constructs a diagram with no colimit.

This question of yours is a good question, because thinking this way (i.e. asking for good categorical properties) has led to lots of theory over the years. I discuss two such cases below, although for your application they will probably not be helpful.


I study model categories, which are categories where one can "do homotopy theory." So $Top$ and $Top_\ast$ are examples. Let $M$ be a model category (feel free to think of $M$ as $Top$) and let $Ho(M)$ be its homotopy category. It turns out that you get arbitrary products and coproducts in $Ho(M)$ for any model category $M$ (see Hovey's book "Model Categories" 1.3.11). Furthermore, you can sometimes compute products on the model category level, as we'll now discuss.

If one wants $F:M\to Ho(M)$ to preserve products, then one wants $M$ to be a monoidal model category, i.e. to have the structure of a monoidal category (for $Top$ this means having a notion of product, which it does) and to satisfy a coherence condition which says that the monoidal products plays nicely with the model structure. It's proven in Mark Hovey's book Model Categories that $Top$ and $Top_\ast$ are monoidal model categories, i.e. have that coherence condition. So $Ho(Top)$ is a monoidal category whose unit is the image under $F$ of the unit in $Top$. In $Top$ the unit is $\ast$ and so it must be $\ast$ again in $Ho(Top)$. Furthermore, this means you can compute products on the model category level, so if someone gives you two objects of $Ho(Top)$ you can find an object $X\in Top$ such that $F(X)$ is the product of in $Ho(Top)$. The same story works with $Top_\ast$, but now the product is the smash product of spaces.

The lack of colimits and limits is remedied in this context by considering homotopy limits and homotopy colimits. The idea here is that if you have a diagram in $Ho(M)$ then that's like a diagram in $M$ where we don't care about replacing objects up to weak equivalence. Given that choice not to care, the homotopy (co)limit is the object completing the diagram, i.e. satisfies a universal property which factors in the ability to replace objects in the diagram by weakly equivalent ones. Here is a good blog post on the subject.


The desire for nice categorical properties was one of the reasons people started to study the stable homotopy category. A great reference here is Margolis's book "Spectra and the Steenrod Algebra". In the introduction he defines the stable homotopy category in several steps and discusses which limits and colimits you are getting at each step. He mentions that even the category of CW complexes does not have good limit structures, and proceeds to move from CW to the Spanier-Whitehead category (which has stable homotopy classes of maps), and then go one farther to $\mathcal{S}$ (the coproduct completion of the subcategory $SW_f$ of finite complexes, a triangulated category). He discusses in chapter 3 that $\mathcal{S}$ has arbitrary coproducts and arbitrary weak colimits. In chapter 5, section 2, he mentions that it has arbitrary products and discusses weak limits.

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Great answer! This is exactly the sort of perspective I was looking for. –  Daniel Miller Feb 2 '13 at 13:04
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A very concrete example of a cospan having no pullback in the homotopy category, which does not require any knowledge of cohomology or Moore spaces, can be found here. It's phrased in terms of the homotopy category of groupoids, but can easily be translated to speak about $K(\pi,1)$s in the homotopy category of spaces instead.

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I don't understand. Maybe the example given in that link shows that a homotopy limit is not always a limit in the homotopy category, but it does not seem to show that there is no limit in the homotopy category. It looks like the limit exists and is a one-point space. –  Tom Goodwillie Feb 2 '13 at 22:29
    
Or rather no: in this case, if I understand right, the homotopy limit is also contractible. –  Tom Goodwillie Feb 2 '13 at 23:11
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@Tom: The shape of the argument is "Suppose there were a limit in the homotopy category. Then ... and so we have a contradiction." In the course of deriving the contradiction, we consider also the homotopy limit of the same diagram, but I don't see how that changes the result we're proving. Is there a mistake somewhere? –  Mike Shulman Feb 3 '13 at 18:41
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@Adam: replace each group $G$ with its classifying space $B G$. –  Mike Shulman Feb 3 '13 at 18:42
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To me the shape of the argument seems perfectly sound argument, although it is given in the context of the homotopy category of categories rather than spaces. If you replace each occurrence of "monoid" with "group" and each group G with BG, then you get an argument for TOP too. The argument goes roughly (1) suppose that there exists a certain pullback Q in the homotopy category (2) The same diagram admits a specific homotopy pull-back P (by a specific calc) (3) Thus there must be a map P --> Q (4) There can be no such map (by another specific calc), hence no such Q. Is there an error in 2? 4? –  Chris Schommer-Pries Feb 4 '13 at 22:41
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The inclusion $\mathbb RP^2\to \mathbb CP^2$ has no kernel in the pointed homotopy category.

Proof:

Suppose $X\to \mathbb RP^2$ is such a kernel. Then maps $S^1\to X$ (in that category) correspond bijectively to maps $S^1\to\mathbb RP^2$ such that the composed map $S^1\to \mathbb CP^2$ is trivial. This means that $\pi_1(X)$ has order $2$.

It follows that there is a map $\mathbb RP^2\to X$ inducing an isomorphism of $\pi_1$. The composed map $\mathbb RP^2\to X\to \mathbb RP^2$ also induces an isomorphism of $\pi_1$, therefore also of $H^1$ with $\mathbb Z/2$ coefficients, therefore (using naturality of cup products) an iso of $H^2$ with $\mathbb Z/2$ coefficients.

But the map $X\to \mathbb RP^2$ must be zero on $H^2$ because the inclusion $\mathbb RP^2\to\mathbb CP^2$ is an isomorphism on $H^2$ and the composed map $X\to \mathbb RP^2 \to \mathbb CP^2$ is zero.

EDIT: Also, the degree two map $S^1\to S^1$ has no cokernel. Briefly, any such cokernel would have to be a retract (up to homotopy) of $\mathbb RP^2$, but that would make it either a point or $\mathbb RP^2$ and neither of these does the job.

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Touch\'{e}:):)! –  Adam Epstein Feb 2 '13 at 15:27
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As Tom points out, many diagrams in the homotopy categories do not even have limits or colimits, so the forgetful functor cannot preserve them!

Meanwhile, I too had difficulty chasing this down, until I found the following remark (M. Mather, Pullbacks in homotopy theory, Can. J. Math. 1976):

"For example, no essential map between Eilenberg-MacLane spaces of different dimensions has a kernel".

My attempt at reconstruction goes as follows. Let $m,n$ be positive integers and let $G,H$ be abelian groups. If $m\neq n$ then any morphism $K(G,m)\rightarrow K(H,n)$ induces the zero morphism $\pi_\ell(K(G,m))\rightarrow\pi_\ell(K(H,n))$ for each $\ell\geq 0$. However, since the set of morphisms $K(G,m)\rightarrow K(H,n)$ is in canonical bijection with $H^n(K(G,m),H)$ which is typically nontrivial, such essential morphisms do exist. To see that such a morphism $K(G,m)\rightarrow K(H,n)$ admits no kernel, note that if $X\rightarrow K(G,m)$ were such a kernel then all the induced maps $\pi_\ell(X)\rightarrow\pi_\ell(K(G,m))$ would be bijections (for $\ell=m$ because $\pi_m(K(H,n))=0$, for $\ell\neq m$ because $\pi_\ell(K(G,m))=0$) so that $X\rightarrow K(G,m)$ would be an isomorphism in the homotopy category, whence $K(G,m)\rightarrow K(H,n)$ must be inessential.

I have been meaning to post my own question regarding whether anyone happens to know a simpler example (or a recipe, or a reference) of a diagram which fails to have co/limits.

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Yes, I do happen to know of a simpler example and posted it right on this thread moments after you posted this answer. It's in the book by Jeffrey Strom, and I linked to a google books preview of the relevant pages –  David White Feb 1 '13 at 23:41
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Well, the first example of an essential map between E-M spaces of different "dimensions" is the inclusion $\mathbb RP^\infty\to \mathbb CP^\infty$. This has no kernel. I guess neither does the inclusion $\mathbb RP^2\to \mathbb CP^2$. –  Tom Goodwillie Feb 2 '13 at 3:32
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True, true. I don't think you got that far when you taught ne in Math 272. –  Adam Epstein Feb 2 '13 at 9:14
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I'm not sure I knew this when I taught you in Math 272. –  Tom Goodwillie Feb 3 '13 at 2:09
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Well, a lot can happen in 29 years. –  Adam Epstein Feb 3 '13 at 8:52
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