How can I show my multiplication for $\pi_n(\Omega X,\ast)$ is unital? Lemma $7.6$, Goerss-Jardine

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I've come up with a candidate construction, however I'm stuck on one last little detail. I'm also bothered by an anxiety that my construction is not the 'right one' and that I've seriously overcomplicated things and missed an easy solution.

The explicit task:

Let $X$ be a nonempty fibrant simplicial set and $\ast\in X_0$ any vertex thereof, $n\in\Bbb N$. If $\Omega X$ is the loop space simplicial set of $X$, based at $\ast$, then classes in $\pi_n(\Omega X,\ast)$ are identifiable with the simplicial homotopy classes of maps: $$\alpha:\Delta^n\times\Delta^1\to X,\,\alpha\circ i=\ast$$Where $i:(\partial\Delta^n\times\Delta^1)\cup(\Delta^n\times\partial\Delta^1)\hookrightarrow\Delta^n\times\Delta^1$, with homotopy taken relative to this boundary $(\partial\Delta^n\times\Delta^1)\cup(\Delta^n\times\partial\Delta^1)$.

Show that there exists a map $\star:\pi_n(\Omega X,\ast)\times\pi_n(\Omega X,\ast)\to\pi_n(\Omega X,\ast)$ with the interchange law satisfied: $[\alpha_1\star\beta_1][\alpha_2\star\beta_2]=([\alpha_1][\alpha_2])\star([\beta_1][\beta_2])$ and with the same unit $e=[\ast]$. Deduce that $\pi_n(\Omega X,\ast)$ is always Abelian.

The only description they give of this map is: "$\alpha\star\beta$... in the $1$-simplex direction", which means nothing to me. 'They' are Goerss-Jardine, this is taken from their book on simplicial homotopy theory.

I have managed to find an operation $\star$ and I've proven it to be well-defined and to satisfy the interchange law. On one hand, that gives me confidence to say my operation is correct. However, I've found it quite hard to prove that $e$ is an identity for $\star$, and my operation does feel weird (my proofs of well-definedness etc. have involved homotopies of homotopies of homotopies, which is not a strategy yet exhibited by Goerss-Jardine and I am suspicious!) so it's very plausible I've overcomplicated things. I'd appreciate comments on the correctness of my approach, as I could not find a solution elsewhere.

The map:

Fix a representative $\alpha$ of some class in $\pi_n(\Omega X,\ast)$. Let $\iota:\Delta^n\times\Delta^1\hookrightarrow\Delta^n\times\Delta^1\times\Delta^1$ be the map $(x,y)\mapsto(x,1,y)$. There is an anodyne extension diagram: Diagram 1

We define $[\alpha]\star[\beta]$ to be $[\psi\circ\iota]$. This makes sense because $\psi\iota$ restricts to $\ast$ on the boundary.

I have proven to my own satisfaction that $\star$ is well-defined w.r.t different choices of representatives and of lifts $\psi$, and I have shown that the interchange law holds. However, as mentioned, I really struggle to show that $[\alpha]\star[\ast]=[\alpha],[\ast]\star[\alpha]=[\alpha]$. I've drawn many mostly futile diagrams in this attempt. One was tantalisingly close to being successful:

Diagram 2

Where $\psi$ is taken from a construction of $[\alpha]\star[\ast]$. The map $\varphi\lambda$ provides a simplicial homotopy $\alpha\simeq\psi\iota$, and this homotopy is relative to $\partial\Delta^n\times\Delta^1$. However, it is not relative to $\Delta^n\times\partial\Delta^1$ in general, unfortunately. I've tried for quite some time to modify this diagram or make similar ones, tweaking the nature of the anodyne extensions, but to no avail - I just don't see it. I might also try to explicitly construct a lift $\psi$ such that $\psi\iota=\alpha$, but one needs to juggle too many conditions to squeeze into one extension, I think.

The question: To what extent is this the 'right' construction of $\star$? How can I demonstrate $\star$ is unital (with unit $e=[\ast]$)?

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As usual, I write this self-answer for the benefit of future readers and so that I might have a formal write-up to take paper notes from later.

JHF's suggested operation was perfect. I still find it very strange that my suggested operation ticks every box but one - what are the odds of it satisfying the interchange law? That gave me false hope for quite some time. I would really appreciate it if a reader shared a way to make my construction work, i.e. shared a proof that it is unital.


This is now a detailed proof of lemma $7.6$.

I leave it to the reader to make the tedious checks that all the maps out of a union make sense (i.e. they agree on the common intersection). All extensions are anodyne, using various lemmae from Goerss-Jardine, so I do not mention this and implicitly assume lifts always exist (which is true for fibrant $X$ and anodyne extensions).

Fix $n\in\Bbb N$ and representatives $\alpha,\beta$ of classes in $\pi_n(\Omega X,\ast)$. There is a map: $$(\Delta^n\times\Lambda^2_1)\cup(\partial\Delta^n\times\Delta^2)\overset{((\alpha,\beta),\ast)}{\longrightarrow}X$$Which has a lift: $$\psi:\Delta^n\times\Delta^2\to X$$We define $\alpha\star\beta$ to be $\psi\circ(1\times\delta_1(-)):\Delta^n\times\Delta^1\hookrightarrow\Delta^n\times\Delta^2\to X$. This map has the property $(\alpha\star\beta)\circ i=\ast:(\partial\Delta^n\times\Delta^1)\cup(\Delta^n\times\partial\Delta^1)\to X$, so $[\alpha\star\beta]\in\pi_n(\Omega X,\ast)$.

To check this is well-defined w.r.t different representatives and lifts:

  • Let $[\alpha]=[\alpha'],[\beta]=[\beta']$ in $\pi_n(\Omega X,\ast)$
  • Let $\psi_1$ witness $\alpha\star\beta$ and $\psi_2$ witness $\alpha'\star\beta'$

There are relative homotopies $h_1:\alpha\simeq\alpha'$ and $h_2:\beta\simeq\beta'$ (in that order, left$\to$right). There is a map: $$(\Delta^n\times\Delta^2\times\partial\Delta^1)\cup(((\partial\Delta^n\times\Delta^2)\cup(\Delta^n\times\Lambda^2_1))\times\Delta^1)\overset{((\psi_1,\psi_2),(\ast,(h_1,h_2)))}{\longrightarrow}X$$

What ordering you place on the faces of $\Lambda^2_1$ is unimportant, it only matters that you choose one ordering for the definition of $\star$ and keep it consistent throughout. By 'ordering', I mean how one should interpret maps such as $(h_1,h_2):\Delta^n\times\Lambda^2_1\times\Delta^1\to X$.

There is a lift: $$\varphi:\Delta^n\times\Delta^2\times\Delta^1\to X$$And we can let $\lambda:=\varphi\circ(1\times\delta_1(-)\times1):\Delta^n\times\Delta^1\times\Delta^1\to X$. We have that $\lambda$ is a relative homotopy $(\alpha\star\beta)\simeq(\alpha'\star\beta')$, as desired. Thus $\star:\pi_n(\Omega X,\ast)\times\pi_n(\Omega X,\ast)\to\pi_n(\Omega X,\ast)$ is a well-defined function.

This $\star$ is unital with unit $e=[\ast]$: since $\star$ is well-defined w.r.t choice of lift $\psi$, we can let $\psi:=\alpha\circ(1\times\sigma_2(-))$ and $\psi:=\alpha\circ(1\times\sigma_0(-))$ to witness $[\alpha]\star e=[\alpha]=e\star[\alpha]$, as these $\psi$ are valid extensions of the respective diagrams (which $\psi$ witnesses what depends on your choice of ordering). This is where JHF's suggestion of using an inner horn $\Lambda^2_1$ becomes crucial!

Finally, we want to verify the interchange law. For this, I recall the definition of ordinary multiplication in the simplicial homotopy groups:

$$[x]\cdot[y]=[d_n\omega]$$Where $x,y$ are $n$-simplices and $\omega$ is an $(n+1)$-simplex with boundary $(\ast,\ast,\cdots,\ast,x,-,y)$.

In this case, we must translate simplices of $\Omega X$ to maps to $X$, and that amounts to: $$[x]\cdot[y]=[\omega\circ(\delta_n(-)\times1)]$$Where $x,y:\Delta^n\times\Delta^1\to X$ represent elements of $\pi_n(\Omega X,\ast)$ and $\omega:\Delta^{n+1}\times\Delta^1\to X$ has: $$\omega\circ(\delta_i\times1)=\begin{cases}\ast&i\le n-2\\x&i=n-1\\y&i=n+1\end{cases}$$

So let $\alpha,\alpha',\beta,\beta'$ be any representatives and let $\omega_1$ witness $[\alpha]\cdot[\alpha']$, $\omega_2$ witness $[\beta]\cdot[\beta']$. Let $\psi_1$ witness $[\alpha]\star[\beta]$ and $\psi_2$ witness $[\alpha']\star[\beta']$. Let $\varphi$ witness the ordinary multiplication $([\alpha]\star[\beta])([\alpha']\star[\beta'])$.

There is a map: $$(\Delta^{n+1}\times\partial\Delta^2)\cup(\Lambda^{n+1}_n\times\Delta^2)\overset{((\omega_1,\varphi,\omega_2),(\ast,\cdots,\ast,\psi_1,-,\psi_2))}{\longrightarrow}X$$with lift: $$\gamma:\Delta^{n+1}\times\Delta^2\to X$$

Define $\lambda:=\gamma\circ(\delta_n(-)\times1):\Delta^n\times\Delta^2\to X$. It can be checked that $\lambda$ witnesses $[\alpha\alpha']\star[\beta\beta']$ (it is an appropriate extension) and that $\lambda\circ(1\times\delta_1(-))=([\alpha]\star[\beta])([\alpha']\star[\beta'])$. This proves: $$([\alpha]\star[\beta])([\alpha']\star[\beta'])=[\alpha\alpha']\star[\beta\beta']$$

As desired. Eckmann-Hilton shows that $\star$ coincides with the usual multiplication and that 'both' are Abelian.