# Nullhomotopicity of $\mathbb{S}/p \to^p \mathbb{S}/p$ for $p=2$ and $p \neq 2$?

For a given spectra $$X$$ we have $$X/p$$ defined as the cofiber $$X \to^p X$$ where the map is basically defined via defining it on the sphere spectrum $$\mathbb{S}$$! To define $$\cdot p$$ on the sphere spectrum we just need to define it on $$\mathbb{S^1}$$ where it's $$z \to z^p$$ in the complex numbers (and then via suspension we define it for higher spheres).

We thus obtain $$\mathbb{S}/p$$ and we can consider $$s_p: \mathbb{S}/p \to^p \mathbb{S}/p$$

I'm watching higher algebra on youtube higher algebra- padic spectra where it is claimed that $$s_p$$ is nullhomotopic for $$p \neq 2$$, but is not nullhomotopic for $$p = 2$$, while $$s_p * s_p$$ should always be nullhomotopic.

I'm trying to verify those claims. Instead of computing the pushout of $$\cdot p$$ on the sphere spectrum which is $$\Sigma^{\infty}(\mathbb{S}^0)$$, since $$\Sigma^{\infty}$$ commutes with colimits we can do the colimit in topological spaces. Because $$\mathbb{S}^0$$ is scarily degenerate I'm going to use $$\mathbb{S}^1$$

We can check that the cofiber $$\mathbb{S}^1 \to^2 \mathbb{S}^1$$ is $$\mathbb{R}P^2$$, and more generally it's the disk where you glued the boundary $$p$$ times instead of just twice, we'll denote this space as $$\mathbb{D}_p$$.

Now let us discuss $$\cdot p$$ on $$\mathbb{D}_p$$ and whether it's nullhomotopic. For $$p=2$$ it certainly is; since $$\cdot 2$$ vanishes on $$\pi_1(\mathbb{D}_2)$$ we can lift $$\cdot 2 : \mathbb{D}_2 \to \mathbb{D}_2$$ to $$\cdot 2 : \mathbb{D}_2 \to \mathbb{S^2}$$ where it lands on the upper hemisphere and so is contractible. More directly, every $$\cdot p$$ on $$\mathbb{D}_p$$ is nullhomotopic via squeezing towards the center of the disk along time.

My proof must be wrong because I was told there is a problem for $$p=2$$, that there one needs to multiply by $$4$$. Where was I wrong? What did I miss?

• I don't think you're in the stable range for your calculation yet.
– JHF
Jun 10 at 18:01
• @JHF but I showed that it's ALREADY nullhomotopic. How can stabiliziation ruin it? Can't I suspension loop space the proof of the nullhomotopy? Jun 11 at 5:45
• I think the point is that multiplication by $p$ map is not defined until you suspend.
– JHF
Jun 13 at 18:07
• @JHF I thought it was defined as I mentioned, how else would you define it after suspenion? Jun 13 at 18:22
• How are you defining the map $\cdot p$ on $\mathbb{D}_p$? Jun 13 at 18:28