Suppose $\mathfrak{M}$ is the category of $S^1$-spectra of simplicial sheaves. I know its sequential homotopy colimits (colimit in $\mathbb{N}$ as usual) coincide with categorial colimits since stable equivalences are preserved under filtered colimits.

But I don't see why they coincide with the homotopy colimits in $\mathrm{Ho}(\mathfrak{M})$, as a triangulated category. That is, $$\mathrm{Hocolim}(E_i):=\mathrm{Cone}(1\text{-}\mathrm{shift}:\oplus E_i\longrightarrow\oplus E_i)$$.

Does anybody know the reason?


There's probably a more classical way to look at this, but here's one way to think about it. Homotopy colimits in $M$ are the same as $\infty$-categorical colimits in the $\infty$-category $M[W^{-1}]$ obtained from $M$ by localizing at the stable equivalences. In particular, sequential homotopy colimits in $M$ are the same as $\infty$-categorical colimits in $M[W^{-1}]$ indexed by $\omega$ -- the poset of natural numbers.

In 1-category theory, there is a formula which expresses any colimit indexed by a category $J$ as a (reflexive) coequalizer of certain coproducts. Similarly, in $\infty$-category theory, there is a formula which expresses any colimit indexed by a simplicial set $J$ (which need not be a quasicategory) as a geometric realization of certain coproducts -- for quasicategories, you can find this in Chapter 4 of HTT. Schematically, it looks like $\varinjlim F = |[n] \mapsto \amalg_{j \in J_n} F(d_1(\dots(d_n(j)))|$. We can get away with the coproduct being over just the nondegenerate $n$-cells. This doesn't really simplify things when $J = \omega$, since (the nerve of) $\omega$ has nondegenerate simplices of arbitrarily large dimension.

However, let $N$ be the same simplicial set which I coincidentally just described answering another question of yours -- its 0-cells are the natural numbers, there is a unique 1-cell from $n$ to $n+1$, and there are no other nondegenerate cells. There is a natural inclusion $N \to \omega$. Using the $\infty$-categorical Quillen's Theorem A (see 3.2 there, or HTT Chapter 4), one easily shows that this inclusion is cofinal (In fact, I think this is explicitly shown somewhere in Chapter 4 of HTT). Therefore, a sequential colimit of a functor $\omega \to \mathcal C$ can be computed by first restricting to get a functor $N \to \mathcal C$ and then evaluating this colimit.

Now when we apply the general formula to our $N$-indexed colimit, since $N$ has no nondegenerate simplices of dimension greater than 1, we only need the first two layers of the simplicial object, so everything simplifies to the coequalizer of two maps $\amalg_{n \in \mathbb N} E_n \rightrightarrows\amalg_{n \in \mathbb N} E_n$, namely the identity and the shift map. In an additive category, this coequalizer can be computed as the (homotopy) cofiber of the difference of these two maps.

Finally, coproducts in $M[W^{-1}]$ are computed as in the homotopy category, and cofibers are computed using the triangulated structure, yielding exactly the formula you described.

If the question is really why the homotopy coequalizer of two maps is the same as the homotopy cofiber of their difference (the last step of the above argument, essentially), then let's go through that. By passing to the opposite category, it will suffice to treat the dual case, and show that a homotopy equalizer of two maps $f,g: X \rightrightarrows Y$ is the same as the fiber of their differences. Homotopy limits, like ordinary limits, are defined representably. That is, a cone in $\mathcal C$ is a limit if and only if it becomes a limit cone after composing with $Hom_{\mathcal C}(C,-): \mathcal C \to Spaces$ for each $C \in \mathcal C$ (where $Hom_{\mathcal C}$ denotes a mappping space). So it will suffice to show this when $\mathcal C = Spaces$, and we assume that $X,Y$ are infinite loop spaces and $f,g$ are infinite loop maps.

Now, a point $(x,\gamma) \in Hoeq(f,g)$ consists of a point $x \in X$ and a path $\gamma$ from $f(x)$ to $g(x)$. A point $(x,\gamma) \in Fib(g-f)$ consists of a point $x \in X$ and a path $\gamma$ from $(g-f)(x)$ to the basepoint 0. A map $Hoeq(f,g) \to Fib(f,g)$ is given by sending $(x,\gamma)$ to $(x,\gamma -f(x))$, where $\gamma-f(x)$ is the path obtained by using the addition on $Y$ to add the constant path at $f(x)$ pointwise to $\gamma$. A map in the other direction is given by sending $(x,\gamma)$ to $(x,\gamma+f(x))$, and these are inverse homotopy equivalenes.

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    $\begingroup$ I don't think I need such a heavy context. I find that the question is to prove that the homotopy coequalizer of two morphisms in $\mathfrak{M}$ is just the cone of their difference in $Ho(\mathfrak{M})$. $\endgroup$ – Nanjun Yang Feb 15 at 9:46
  • $\begingroup$ I’d be curious how you reduce to that statement. It’s quite elementary to show that coequalizers are cofibers of differences. $\endgroup$ – Tim Campion Feb 15 at 15:15
  • $\begingroup$ Maybe I’ll also point out that this infinity categorical argument is exactly the same argument one would make 1-categorically, just lifted to the higher context. $\endgroup$ – Tim Campion Feb 15 at 15:23
  • $\begingroup$ How about the homotopy coequalizer of two identical maps? $\endgroup$ – Nanjun Yang Feb 16 at 2:36
  • $\begingroup$ Yup, the homotopy coequalizer of two identical maps $X \rightrightarrows Y$ is the cofiber of the zero map $X \xrightarrow 0 Y$, which is none other than $Y \oplus \Sigma X$. $\endgroup$ – Tim Campion Feb 16 at 2:37

First, an elementary manipulation of homotopy colimits shows that the sequential homotopy colimit can be replaced by the homotopy coequalizer of the identity map and the shift map on the coproduct of all objects. The latter homotopy coequalizer can be replaced by the homotopy coequalizer of the zero map and the difference of the identity and shift map.

Secondly, by definition of a triangulated category extracted from a stable model category, the distinguished triangles are defined precisely as the homotopy coequalizers of the zero map and the given map.

Combined with the fact that homotopy coproducts in a stable model category coincide with coproducts in its underlying triangulated category, we obtain the desired result.

  • $\begingroup$ The point is that $Ho(\mathfrak{M})$ is an additive category but $\mathfrak{M}$ is NOT, so you could't talk about 'difference of maps in $\mathfrak{M}$'. $\endgroup$ – Nanjun Yang Feb 16 at 0:44
  • $\begingroup$ @NanjunYang: The difference of maps mentioned in the first paragraph is not induced by any kind of additive structure. Rather, it is induced by the fact that Map(X,Y) is weakly equivalent to Map(X,ΩΣY), which is a group object in spaces, which allows us to compute the difference of two maps. $\endgroup$ – Dmitri Pavlov Feb 16 at 6:11

I've worked out.

For any two morphisms $f,g:A\to B$ in $\mathfrak{M}$, we have $$Cone(f-g)=Cone(((id_A,f),(id_A,g)):A\oplus A\to A\oplus B)$$ by elementary transformations of matrices.

We prove that latter is the homotopy pushout of the diagram $$A\xleftarrow{id+id}A\vee A\xrightarrow{f+g}B$$ , which is just $Hocoeq(f,g)$. So just assume we have a general pushout diagram $\require{AMScd}$ \begin{CD} W @>t>> V\\ @V s V V @VV a V\\ U @>>b> D \end{CD} in $\mathfrak{M}$, such that $W$ is cofibrant and $t$ is a cofibration. So $Cone(t)=Cone(b)=D/U=V/W$ by definition and left properness of $\mathfrak{M}$. By Lemma 1.4.3 in 'Neeman Triangulated Category', we have $$Cone(W\xrightarrow{(s,t)}U\oplus V)=D$$ by possibly modifying $a$. Thus we have proved the claim. Hence we have proved $$Cone(f-g)=Hocoeq(f,g)$$.


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