# Suspensions are H-cogroup objects

Do you know any reference where you have a formal justification for the following statement that appears in nLab?

https://ncatlab.org/nlab/show/suspensions+are+H-cogroup+objects

"Let $\mathcal{C}$ be an $(∞,1)$-category with finite $(∞,1)$-colimits and with a zero object. Write $\Sigma \colon X \mapsto 0 \underset{X}{\coprod} 0$ for the reduced suspension functor.

Then the pinch map

$$\Sigma X \simeq 0 \underset{X}{\sqcup} 0 \simeq 0 \underset{X}{\sqcup} X \underset{X}{\sqcup} 0 \longrightarrow 0 \underset{X}{\sqcup} 0 \underset{X}{\sqcup} 0 \simeq \Sigma X \coprod \Sigma X$$

exhibits a cogroup structure on the image of $\Sigma X$ in the homotopy category of an $(∞,1)$-category homotopy category $Ho(\mathcal{C})$".

• What do you not understand on the nlab page? It might help to see the classical example, where $C$ is the ∞-category of pointed spaces. There the cogroup structure is given by "pinching" the suspension at the equator (note that you can identify $0\amalg_X X$ and $X\amalg_X 0$ with the two cones forming the suspension). – Denis Nardin May 7 '18 at 17:05
• See also the discussion here for more details – Denis Nardin May 7 '18 at 17:07
• There the cogroup structure is given by "pinching" satisfying the diagrams for co-associativity, co-unit, and co-inverse. Is this trivial? – Andres Felipe Ramírez May 7 '18 at 17:33
• Well, I don't know about trivial, but it is about as hard as proving that concatenation of loops gives a group structure (in fact it is pretty much an equivalent statement, since $[ΣX,Y]=\pi_1Map(X,Y)$) – Denis Nardin May 7 '18 at 17:35
• As it proves that the that concatenation of loops gives a group structure? (in the context of the (∞,1) -categories) – Andres Felipe Ramírez May 7 '18 at 17:54

Ok, let me try to give you a proof of something that is a lot stronger than what you asked for, but which hopefully is a bit more natural. I am basically going to smother the problem under the abstract nonsense, so I recommend you look at special cases of this proof, e.g. when $X=S^n$ in pointed spaces, in order to get intuition.

Recall that an associative monoid in an ∞-category $D$ with all finite products is a functor $$M:\Delta^{op}\to D$$ such that for all $[n]\in \Delta^{op}$ the map $$M([n])\xrightarrow{\prod M(e_i)} \prod_{1\le i\le n} M([1])$$ is an equivalence, where $e_i:[1]\to [n]$ is the map picking out the edge $\{i-1,i\}$ (this is known as the Segal condition). Hence, an associative comonoid in an ∞-category with all finite coproducts $C$ is a functor $$M:\Delta\to C$$ such that $$\coprod_{1\le i\le n}M([1])\xrightarrow{\coprod M(e_i)} M([n])$$ is an equivalence. Moreover, by postcomposing with the map $C\to hC$, we see that an associative comonoid in $C$ induces an associative comonoid in $hC$.

What I will prove is that, if $C$ is a pointed ∞-category with finite colimits, $\Sigma X$ is an associative comonoid for all $X$ (the same proof, maybe more recognizably, will prove that $\Omega X$ is a group object for all $X$).

Let us consider the category $\Delta_+$ of all finite (possibly empty!) totally ordered sets. This contains two full subcategories we are going to use: $\Delta$ (the category of finite nonempty totally ordered sets) and $\Delta^1$ (the full subcategory spanned by $[0]$ and by the empty set $\varnothing$).

We can construct a functor $M_0:\Delta^1\to C$ sending $[0]$ to $0$ and $\varnothing$ to $X$. Let $M_+$ be the left Kan extension of $M_0$ to $\Delta_+$.

Lemma The restriction of $M_+$ to $\Delta$ is an associative comonoid $M$ such that $M([1])\cong \Sigma X$.

Proof. Unwrapping the standard formula for left Kan extensions we obtain $$M([n])= \mathrm{colim}_{i\in \Delta^1_{[n]/}} M_0(1) = 0\amalg_X 0\amalg_X \cdots \amalg_X 0$$ where in the formula there are $n+1$ 0s. In particular $M([1])=\Sigma X$. For the sake of clarity of notation let me prove the Segal condition only for $n=2$. Then we need to prove the map $$(0\amalg_X 0)\amalg (0\amalg_X 0) \to 0\amalg_X 0 \amalg_X 0$$ induced by the inclusions of the corresponding summands is an equivalence. But this is simply the associativity of the pushout. $\square$

Finally let me say a few words about how to prove $\Sigma X$ has the structure of a cogroup and not simply a comonoid. If we let $$m:M([1])\to M([2])\cong M([1])\amalg M(1)$$ be the map induced by the edge $\{0,2\}$ (the "comultiplication"), the condition of being a cogroup is equivalent as asking the map $$M([1])\amalg M([1])\xrightarrow{m\amalg i_1} M([1])\amalg M([1])$$ is an equivalence. To check this is the case for the $M$ I constructed above is left as an exercise for the reader.

I do not know if this answer is the one you are looking for. I apologize if it is not. The classical statement--for topological spaces, not for an arbitrary infty-category--is textbook stuff, specifically section 1.6 of Spanier's textbook. Spanier's proof there consists of the sentence "The proofs of the following thoerems are dual to the proofs of the corresponding statements about H-groups[...]and are omitted," and while Spanier's proofs for H-groups are more topological in nature, it is not difficult to see how to make them formal and categorical, and consequently how to make them work in any pointed model category--or any finitely complete pointed infty-category, I guess.