let $hTop_*$ denote the homotopy category of pointed spaces. I believe that it has no pushouts, in general. the reason is that you can't expect the involved homotopies to be compatible. can anyone give an explicit example, with proof? I know that homotopy colimits are related to this, but they don't seem to be categorical colimits, so I don't think that they fit here.

especially I'm interested in the following special case: let $G= \langle X | R \rangle$ a presentation of a group and consider the resulting map $\omega : \vee_{r \in R} S^1 \to \vee_{x \in X} S^1$. does the cokernel of $\omega$ exist in $hTop_*$? in $Top_*$, the cokernel is just consider the 2-dimensional CW-complex $Q$, which is optained from $\vee_{x \in X} S^1$ via the attaching map $\omega$. now if $f : \vee_{x \in X} S^1 \to T$ is a pointed map such that $f \omega$ is nullhomotopic, it is easy to see that it extends to a map $\overline{f} : Q \to T$. but I think that we cannot expect that $\overline{f}, \overline{g}$ are homotopic, when $f,g$ are homotopic: the homotopies between $f$ and $g$ don't have to be compatible. can you give an example for that? probably it already works for $\omega : S^1 \to S^1, z \mapsto z^2$, thus $Q = \mathbb{R} P^2$.

anyway, this only would show that $Q$ is not the cokernel in the category $hTop_*$. the proof, that the cokernel does not exist at all, will be even more difficult and I don't know how to approach it.

you may also replace the category by $hCW_*$ (CW-complexes), $hCG_*$ (compacty generated spaces) etc., if it's useful.


Your example (the "cokernel" of the multiplication by 2 map) also works.

Consider the diagram $S^1 \leftarrow S^1 \rightarrow D^2$ in the based homotopy category of CW-complexes, where the left-hand map is multiplication by 2. Suppose it had a pushout $X$ in the homotopy category. Then for any $Y$, $[X,Y]$ is isomorphic to the set of 2-torsion elements in $\pi_1(Y)$.

Taking $Y = S^0$, we find $X$ is connected.

Taking $Y = K(\pi,1)$, we find that $\pi_1(X)$ must be isomorphic to $\mathbb{Z}/2$. This means that there is a map from ${\mathbb{RP}^2}$ to $X$ inducing an isomorphism on $\pi_1$, and that there is a map $X \to K(\mathbb{Z}/2,1)$ that also induces an isomorphism on $\pi_1$.

Net result, we get a composite sequence of maps $\mathbb{RP}^2 \to X \to \mathbb{RP}^\infty \to \mathbb{CP}^\infty$. The final space is simply connected, so the map from $X$ would be nullhomotopic and hence so would the map from $\mathbb{RP}^2$.

However, the composite of the first two maps is an isomorphism on $\pi_1$, hence on $H_1$. Looking at induced maps on the second cohomology group $H^2$, we get the sequence of maps: $$\mathbb{Z}/2 \leftarrow H^2(X) \leftarrow \mathbb{Z}/2 \leftarrow \mathbb{Z}$$ The rightmost map is surjective, the composite of the two leftmost maps is an isomorphism by the universal coefficient theorem, and the composite of the two rightmost maps is supposed to be nullhomotopic and hence zero. Contradiction.

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    $\begingroup$ thank you :). by the way, this shows that $\pi_1 : hTop_* \to Grp$ has no left adjoint. $\endgroup$ – Martin Brandenburg Jan 1 '10 at 21:20

A standard example is the following: The push out of $D^2\hookleftarrow S^1 \hookrightarrow D^2$, where the arrows are boundary inclusions, seems to be $S^2$. Now in $hTop_*$ the two 2-discs $D^2$ represent the same object as the point $*$, so the diagram $*\hookleftarrow S^1 \hookrightarrow *$ is the same in $hTop$ and its push out would be the homotopy class of a point, but $S^2\not\sim *$. Homotopy colimits are not categorical, but make use of the model structure and tell you how to build "homotopy-correct" colimits in your model structure. I think this is one of the main reasons why one is happy to have not only the existence of a certain homotopy category, but also the model structure on the localizer. So you can do computations in the model category with the help of the model structure and watch the results in the homotopy category.

  • $\begingroup$ Is the latex code displayed correct? On my screen its displayed wrong as long as I click on edit. Then, in the preview, everything is fine. Strange?! $\endgroup$ – user2146 Jan 1 '10 at 14:03
  • $\begingroup$ I added some backquotes to keep the *'s from messing things up. $\endgroup$ – Tyler Lawson Jan 1 '10 at 14:25
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    $\begingroup$ I don't think this is what the asker is looking for, which is an example of a diagram in hTop with no pushout at all in hTop. $\endgroup$ – Reid Barton Jan 1 '10 at 14:29
  • $\begingroup$ philip: I know this example ($D^1$ should be $D^2$, by the way), but it has nothing to do with my question. please read my question carefully. $\endgroup$ – Martin Brandenburg Jan 1 '10 at 14:34
  • $\begingroup$ $D^{2}$, sure. Thanks. I am still surprised that my answer has nothing to do with your question. However I had problems with the second paragraph of your question, so I decided to ignore that part. I still wonder why the cokernel of $\omega$ is a 2-dim CW complex. $\endgroup$ – user2146 Jan 1 '10 at 17:43

I'll work in CW-complexes.

Consider the diagram $S^1_+ \leftarrow *_+ \rightarrow S^1_+$, where $X_+$ denotes $X$ with a disjoint basepoint added. Homotopy classes of maps $*_+ \to Y$ are path components of $Y$, and homotopy classes of maps $S^1_+ \to Y$ are a choice of path component and a conjugacy class of element of $\pi_1$ of that path component. Therefore, if this diagram has a pushout in the homotopy category of based spaces it (co?)represents the functor sending $Y$ to a choice of path component and a pair of conjugacy classes in the same path component.

Suppose we had a representing object $X$. Considering $[X,S^0]$, we find $X$ has only two path components. So $X = X_0 \coprod X_1$ where $X_0$ is the basepoint component. Each component can, up to homotopy equivalence, be constructed as a CW-complex with one zero-cell, some family of 1-cells, and some family of 2-cells.

Consider $[X, K(\pi,1)_+]$ for $\pi$ a group. A cell description of $X$ gives rise a description of this functor: an element of $[X, K(\pi,1)]_+$ is either trivial (if $X_1$ maps to the basepoint) or is a conjugacy class of homorphism $\pi_1(X_1) \to \pi$, because the maps on $X_1$ are not restricted to basepoint-preserving homotopies.

So it suffices to show that there are no groups $G$ so that conjugacy classes of homomorphism $G \to \pi$ are in natural bijective correspondence with pairs of conjugacy classes of elements of $\pi$.

EDIT: Fixed up the following argument.

Given such a group $G$, the identity map determines a pair of conjugacy classes $[x], [y]$ in $G$, and choosing any representatives $x$ and $y$ determines a group homomorphism $F_2 \to G$. Conversely, the conjugacy classes of the generators of $F_2$ determine a map $G \to F_2$ splitting this map up to conjugacy. This would imply that the natural tranformation sending simultaneous conjugacy classes of pairs to pairs of conjugacy classes is an inclusion, which is false, e.g. in the symmetric group on 3 letters there are 9 pairs of conjugacy classes and (assuming I counted correctly) 11 simultaneous conjugacy classes of pairs.

  • $\begingroup$ Are you sure about this last argument? Let’s call $f:F_2\to G$ and $g:G\to F_2$. It should be clear that $[f\circ g]=[\mathrm{id}]$ by your argument. What you now seem to consider is the map $-\circ[g]:\mathrm{Hom}(F_2,\pi)/\text{conj}\to \mathrm{Hom}(G,\pi)/\text{conj}$, but this should be split epic, not split monic. So you showed that “conjugacy classes of pairs” surjects onto “pairs of conjugacy classes” – which is true. Or did I make a mistake? $\endgroup$ – FKranhold Apr 25 '19 at 20:07
  • $\begingroup$ @FKranhold You are correct and I apologize for the mistake -- $G$ is a retract of $F_2$ and not vice versa. So I would have to give a better argument about the structure of the group $G$ to complete this argument. $\endgroup$ – Tyler Lawson May 1 '19 at 7:07

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