Is there a 'coherence' theorem for simplicial sets? On uniquely lifting natural transformations of $n$-truncations to full natural transformations

Question

$\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{^{\mathsf{op}}}\newcommand{\sk}{\operatorname{sk}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\cosk}{\operatorname{cosk}}\newcommand{\nat}{\mathsf{Nat}}\newcommand{\graph}{\mathsf{Graph}}\newcommand{gr}{\operatorname{Gr}}$I have been studying 'simplicial stuff' only for a few days, but I've repeatedly come across the same problem. The basic structure of the problem is this:

Given $n\in\Bbb N_0$, two simplicial sets $X,Y$ and some map $\varphi:X\to Y$ which is natural in the first $n$ components, we need to show $\varphi$ is natural in all components given that $X$ has dimension $\le n$. This is very easy when $n=0$, but in higher dimensions I find it confusing - but it is essential for the basic theory of the (co)skeleton and I expect to see the same problem arise in other situations.

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Motivational examples:

• Showing that the definition of the $n$-skeleton as: $$\sk_n(X_m):=\{\sigma\in X_m:\exists k\le n,\,\exists f\in\Delta(m,k),\,\sigma\in X_f(X_k)\}$$Coincides with the definition $n$-skeleton of a simplicial set via left Kan extension. Specifically, it must be shown that a family of functions $q_t:X_k\to Y$, where $Y$ is some set and $t\in\Delta(m,k)$ any arrow, that satisfy $q_{ft}=q_t\circ X_f$ for any $f\in\Delta(k,k')$ (where $k,k'\le n$) must also satisfy: $$q_t(\tau)=q_{t'}(\tau')$$Whenever $X_t(\tau)=X_{t'}(\tau')$.
• Given the definition of $\cosk_n(X_m)$ as: $$\cosk_n(X_m)\cong\nat(\tr_n\Delta^m,\tr_n X)$$We might want to verify: $$\cosk_n(X_m)\cong\nat(\sk_n\Delta^m,X)$$Which amounts to the following: given a family of maps $\varphi_k:\Delta^m_k=\sk_n(\Delta^m_k)\to X_k$, $k\le n$ that is natural in the first $n$ coordinates, and given the definition of $\varphi$ on $k>n$ as: every element of $\sk_n(\Delta^m_k)$ is an arrow $f:k\to m$ that factors as $f=k\overset{g}{\longrightarrow}\nu\overset{h}{\longrightarrow}m$ where $\nu\le n$, so put: $\varphi_k(f):=X_g(\varphi_\nu(h))$. Is this well-defined?
• In a suitable category of graphs $\graph$, there is a functor $\gr:[\Delta\op,\set]\to\graph$ assigning to every simplicial set its graph with $0$-simplices for vertices and nondegenerate $1$-simplices for edges. When $X,Y$ are two simplicial sets, $X$ having dimension $\le1$, we want to show: $$\graph(\gr(X),\gr(Y))\cong\nat(X,Y)$$Which amounts to showing that two maps $\psi_{0,1}:X_{0,1}\to Y_{0,1}$, which are natural, extend to a natural transformation $\psi:X\implies Y$ uniquely. As $X$ is $\le1$-dimensional, an element $\sigma\in X_m$ is expressible as $X_t(\tau)$ for some $\tau\in X_{0,1}$ and we might hope to put $\psi(\sigma):=Y_t\psi(\tau)$ to preserve naturality. But - is this well defined? What if $\sigma=X_{t'}(\tau')$ for other pair $t',\tau'$?

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I initially thought I could argue as follows (taking the notation of the introduction) there is a decomposition of a simplex $\sigma\in X_m$ as $X_\alpha(\omega)$ where $\alpha:m\to n$ surjects.

"Because $\alpha$ surjects, any other $t:m\to k$, $k\le n$, factors as $t=\beta\alpha$. So, if $X_t(\tau)=\sigma$, then $X_\alpha(X_\beta(\tau))=X_\alpha(\omega)$ and because $\alpha$ surjects, $X_\alpha$ injects, so $\omega=X_\beta(\tau)$ and we can derive: $$Y_\alpha\varphi(\omega)=Y_\alpha\varphi(X_\beta(\tau))=Y_\alpha Y_\beta\varphi(\tau)=Y_t\varphi(\tau)$$So the two agree - the definition of $\varphi(\sigma)$ as $Y_t\varphi(\tau)$ for any $t,\tau$ is well-defined."

But this has the fatal flaw that $t=\beta\alpha$ isn't necessarily true. How can we resolve this problem? Why should: $$Y_t\varphi(\tau)=Y_{t'}\varphi(\tau')$$For a map $\varphi$ natural in low-dimensional coordinates, if $X_t(\tau)=X_{t'}(\tau')$?

P.S. I phrased the title as such because this seems like a general principle that might be proven as a subcase of a general theorem about the simplicial sets. It reads like: 'all paths commute' where a path is a decomposition $\sigma=X_t(\tau)$ and the commutativity reads as $Y_t\varphi(\tau)=Y_{t'}\varphi(\tau')$.

Answer 1

$\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{^{\mathsf{op}}}\newcommand{\sk}{\operatorname{sk}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\cosk}{\operatorname{cosk}}\newcommand{\nat}{\mathsf{Nat}}\newcommand{\graph}{\mathsf{Graph}}\newcommand{gr}{\operatorname{Gr}}\newcommand{\lan}{\operatorname{Lan}}$Yet another self-answer - I hope it's right this time! The result is actually not so hard to show. An important prerequisite for the definition of 'dimension' and a certain lemma is given here.

Lemma:

Let $S$ be any simplicial set and $m\in\Bbb N_0$, $\sigma\in S_m$ an arbitrary $m$-simplex. There is a canonical representation: $$\tag{1}\sigma=X_\alpha(\omega)$$Where $\lambda\in\Bbb N_0$ and $\alpha\in\Delta(m,\lambda)$ is a surjection and $\omega\in X_\lambda$ is a nondegenerate simplex. This representation $(1)$ is unique among surjective $\alpha$ and nondegenerate $\omega$. Moreover, $\lambda$ is the minimal integer $j$ for which $\sigma$ is the image of a $j$-simplex.

Precise theorem statement (this is more or less equivalent to the statement that the $n$-skeleton is the composition $\lan_{\iota:\Delta\op_{\le n}\to\Delta\op}\circ\tr_n$):

Let $n\in\Bbb N_0$ and $X,Y$ any two simplicial sets. Suppose we are given a family of functions: $$\left(\varphi_i:X_i\longrightarrow Y_i\right)_{i=0}^n$$Which assemble into a natural transformation $\varphi:\tr_n X\implies\tr_n Y$.

Fix an arbitrary $m\in\Bbb N_0$ and an arbitrary $m$-simplex $\sigma\in X_m$, where $\sigma$ is in the image of a $(k\le n)$-simplex. Among all $0\le k\le n$ and all pairs $(t,\tau)$ where $t\in\Delta(m,k)$, $\tau\in X_k$ have $X_t(\tau)=\sigma$, we have: $$\tag{2}(Y_t\circ\varphi_k)(\tau)=(Y_\alpha\circ \varphi_\lambda)(\omega)$$Where $\sigma=X_\alpha(\omega)$ are as in $(1)$. $\lambda\le n$ is forced by minimality, so $\varphi_\lambda$ makes sense.

As a consequence, when $X$ has dimension $\le n$, there is a unique natural transformation $\psi:X\implies Y$ with components $\psi_i=\varphi_i$ for $0\le i\le n$, given by: $$\tag{3}\psi_m(\sigma)=(Y_t\circ\varphi_k)(\tau)$$For any $k,t,\tau$ as above. This is well defined by the equivalence $(2)$ and the fact that a representation as in $(1)$, with $\lambda=k\le n$ and $t=\alpha$, always exists. Naturality follows from this well-definedness and uniqueness follows from that fact that, if $\psi$ is natural, then: $$\psi_m(\sigma)=\psi_m(X_\alpha(\omega))=(Y_\alpha\circ\psi_\lambda)(\omega)=(Y_\alpha\circ\varphi_\lambda)(\omega)$$So the value of $\psi_m$ on every $m$-simplex is forced.

Proof - by induction:

It just suffices to demonstrate the equivalence $(2)$. Let all symbols be as in the theorem statement.

When $n=0$, the result is trivial. For $\lambda\le n=0$ so $\lambda=0$ and $(2)$ holds precisely because there is one and only one $0\le k\le 0$ and map $t\in\Delta(m,k=0)$, that is, $t=\alpha$. So suppose now $n\in\Bbb N$ and the theorem is proven for all integer values lower than $n$. Fix $m,\sigma,\alpha,\lambda,\omega$ as in the theorem statement, and fix some $k,t,\tau$ too (which are taken arbitrarily).

Case $1$: $\lambda<n$.

If $k<n$, then the equation $(2)$ holds by induction, so suppose $k=n$.

Case $1i)$: $\tau$ is nondegenerate. Then from the lemma, $t$ cannot be surjective. We can factorise $t=m\overset{\delta}{\twoheadrightarrow}a\overset{\gamma}{\hookrightarrow}n$ where $\delta$ injects and $\gamma$ surjects - as $t$ is not surjective, $a<n$ is forced. By induction and the low-dimensional naturality of $\varphi$: $$(Y_\alpha\circ\varphi_\lambda)(\omega)=(Y_\delta\circ\varphi_a)(X_\gamma(\tau))=(Y_\delta Y_\gamma\circ\varphi_n)(\tau)=(Y_t\circ\varphi_k)(\tau)$$Which is exactly what was to be shown.

Case $1ii)$: $\tau$ is degenerate. By the lemma, $\tau=X_{\alpha'}(\omega')$ where $\alpha'\in\Delta(n,a)$ and $\omega'\in X_a$ is nondegenerate, $a\le n$. As $\alpha'$ is an arrow in $\Delta_{\le n}$ and $\varphi$ is natural among such arrows, we can say: $$(Y_{\alpha't}\circ\varphi_a)(\omega')=(Y_t\circ\varphi_n)(\tau)$$That just means we can reduce to case $1i)$ - applying it to $\omega'$ - which was already shown to hold.

Case $\lambda=n$:

Then $k=n$ is forced. If $t=\alpha$, then $\tau=\omega$ is forced by the uniqueness of the lemma and we are done. If $\tau$ is nondegenerate, and $t\neq\alpha$, then $t$ cannot surject and $t=m\overset{\delta}{\twoheadrightarrow}a\overset{\gamma}{\hookrightarrow}n$ for some $a<n$. But then: $$\sigma=X_\delta(X_\gamma(\tau))$$Where $X_\gamma(\tau)$ is an $(a<n)$-simplex, and this contradicts the minimality of $\lambda$. If $\tau$ is degenerate, then $\tau=X_\epsilon(\omega')$ for some $\epsilon:n\to a$, $\omega'\in X_a$ and $a<n$. Again, $\sigma=X_t(\tau)=X_{\epsilon t}(\omega')$ where $\omega'$ is an $(a<n)$-simple, contradicting the uniqueness of the lemma. Therefore $t=\alpha,\tau=\omega$ is forced and we are done.

Why does this cover my examples? Well, this theorem can be translated to: $$\nat(\sk_nX,Y)\cong\nat(\tr_n X,\tr_n Y)$$

As a corollary, the method of proof handles my first example. But more abstractly, the first example is about showing $\sk_n$ to be the composite $\lan_n\circ\tr_n$. This means to say that natural transformations $\sk_n X\implies Y$ are naturally equivalent (via the obvious unit of the left Kan extension) to natural transformations $\tr_n X\implies\tr_n Y$, but that is exactly the theorem.

The second example is exactly the theorem with $Y=X$ and $X=\delta^m$.

The third example is exactly the theorem with $n=1$. Symbolically: $$\nat(X,Y)=\nat(\sk_1(X),Y)$$And: $$\graph(\gr(X),\gr(Y))\cong\nat(\tr_1 X,\tr_1 Y)$$Is easy to show. So put two and two together!