# How to define 0-sphere in a category with zero object?

The 0-sphere $$S^0$$ is the coproduct of two points, $$S^0 \simeq \ast \coprod \ast$$

How to define 0-sphere in a category with zero object?

Let $$\mathcal{C}$$ be a category. A cylinder, $$\mathbf{I}$$, on $$\mathcal{C}$$ is a functor (cylinder functor)

$$I:\mathcal{C} \longrightarrow \mathcal{C}$$

together with three natural transformations

$$e^{0}: 1_{\mathcal{C}} \Longrightarrow I , e^{1}: 1_{\mathcal{C}} \Longrightarrow I, \sigma: I \Longrightarrow 1_{\mathcal{C,}}$$

such that $$\sigma e^{0}= \sigma e^{1}= 1,$$ with $$1: 1_{\mathcal{C}} \Longrightarrow 1_{\mathcal{C}}.$$

Definition 1. Let $$\mathcal{C}$$ be a category with a terminal object other than the initial one. The 0-sphere $$S^0$$ is the pushout in $$\mathcal{C}:$$

$$\require{AMScd}$$ $$\begin{CD} \emptyset @>>>\ast \\ @V V V @VV V\\ \ast @>> > S^0 :=\ast \coprod \ast \end{CD}$$

Definition 2. Let $$\mathcal{C}$$ be a category with a terminal object other than the initial one.

1. The cone $$C^0(X)$$ is the pushout in $$\mathcal{C}:$$

$$\require{AMScd}$$ $$\begin{CD} X @>>^{e^{0}_X}> I(X) \\ @V V V @VV^{\pi_{0}} V\\ \ast @>> > C^0(X):=\ast \underset{X}{\sqcup} I(X) \end{CD}$$

2. The cone $$C^1(X)$$ is the pushout in $$\mathcal{C}:$$

$$\require{AMScd}$$ $$\begin{CD} X @>>^{e^{1}_X}> I(X) \\ @V V V @VV^{\pi_{1}} V\\ \ast @>> > C^1(X):=I(X) \underset{X}{\sqcup} \ast \end{CD}$$

3. The Suspension $$\Sigma(X)$$ is the pushout in $$\mathcal{C}:$$

$$\require{AMScd}$$ $$\begin{CD} X @>>^{\pi_{1}\circ e^{0}_X}> C^1(X) \\ @V ^{\pi_{0}\circ e^{1}_X} V V @VV V\\ C^0(X) @>> > \Sigma(X):=C^0(X)\underset{X}{\sqcup}C^1(X) \end{CD}$$

4. The $$n$$-sphere $$S^n:=\Sigma(S^0)$$ for $$n>0$$.

• The question is not clear. What counts as "$n$-sphere" and "$0$-sphere" in a given category depends on how you're treating geometry in that category. Are you thinking of homotopy and model categories, or perhaps more along the lines of homological algebra, or something else? Oct 14 '18 at 8:30
• All terminal objects are isomorphic. If you need to "separate" them (whatever that means), then you will have to od so using mechanisms which are not invariant under isomorphism, at which point you might as well observe that they are not equal. That will separate them, Oct 14 '18 at 16:14
• There isn't enough context for a sensible answer. You should provide us with more information on what you're doing and what extra properties your category has. Oct 14 '18 at 16:36
• A category with zero object is canonically equivalent to its category of pointed objects, so you way as well assume you are dealing with a category of pointed objects. Then you could ask if the smash product is defined, and if so, is there is unit object. That object will be the zero sphere. Oct 14 '18 at 21:09
• @AndresFelipeRamírez I see you posted this on MO and on MSE with only a few hours' separation. Earlier question here: math.stackexchange.com/questions/2954461/…. This causes people to duplicate each other's efforts, and should be avoided. Oct 14 '18 at 22:19

$$\require{AMScd}$$I guess every sensible definition of "sphere" in a pointed category $$\mathcal C$$ depends on a notion of homotopy pushout, or homotopy cofiber in $$\mathcal C$$; in order for this to be defined you might want to impose at least the structure of a cofibration category on $$\mathcal C$$. Once you have defined it, the zero map $$S^0 \to *$$ can be factored as $$S^0 \to CS^0 \to *$$, where the first arrow is a cofibration, and the suspension of $$S^0$$ is defined by the pushout $$\begin{CD} S^0 @>>> * \\ @VVV @VVV \\ CS^0 @>>> \Sigma S^0 \end{CD}$$ (it is the homotopy pushout of the diagram $$* \leftarrow S^0\to *$$, because you replaced one of its legs with a cofibration, such that $$CS^0\simeq *$$). Of course, having in mind $$\mathcal C = \bf Top$$, $$S^0 \to CS^0$$ s the inclusion of $$S^0$$ in its mapping cone, and its cofiber is $$S^1$$; so, you might want to define $$\Sigma S^0 := S^1$$, and $$\Sigma^n S^0 := S^n$$. Note that
• the factorization of zero maps is not unique; but there exists a weak equivalence between any two $$CS^0$$ and $$C'S^0$$.
• you really need that pushout to be homotopic; the colimit of $$* \leftarrow S^0\to *$$ is always the zero object.
• I feel a bit frown for this downvote, I was simply trying to interpret what the OP is asking. I see the discussion derailed in the comments, though. I didn't specify that in order for $S^0$ to exist in the form $*\amalg *$ you must admit at least this coproduct in $\mathcal C$, but to my eye this can be solved asking for an edit... Oct 14 '18 at 19:17
• @FoscoLoregian But what does $*\sqcup *$ mean in a pointed category? Oct 14 '18 at 22:15