Simplicial Sets, Horns, and Kan Fibrations

Proof. Each identity is checked on an argument $l$.

For equation 190.1 with : the left side omits $i$ and then $j$ from the values, so its image is $[n]\setminus \{i,j\}$, and it sends $l$ to $l$ for , to $l+1$ for , and to $l+2$ for $l\ge j-1$. The right side first omits $j-1$ and then $i$; since , the value $j-1$ of the inner map is shifted to $j$ by ${\delta }^i$, and one computes the same three cases: $l\mapsto l$ for , $l\mapsto l+1$ for , and $l\mapsto l+2$ for $l\ge j-1$.

For equation 190.2 with $i\le j$: both sides send $l$ to $l$ for $l\le i$, to $l-1$ for , and to $l-2$ for . On the left, the inner ${\sigma }^i$ subtracts one above $i$ and the outer ${\sigma }^j$ subtracts one above $j$; on the right, the inner ${\sigma }^{j+1}$ subtracts one above $j+1$ and the outer ${\sigma }^i$ subtracts one above $i$. Substituting the three ranges of $l$ gives the same values.

For equation 190.3 take $i=j$: ${\sigma }^j({\delta }^j(l))$ is ${\sigma }^j(l)=l$ for and ${\sigma }^j(l+1)=l$ for $l\ge j$, so the composite is the identity; the case $i=j+1$ is the same calculation with the value $j+1$ omitted and then collapsed. For the composite omits $i$ and collapses at $j$, and those operations act on disjoint parts of the range, so the composite equals first collapsing at $j-1$ — the index of $j$ before insertion — and then omitting $i$. For the same reasoning applies with the roles reversed. ◻

Proof. Let be the elements of $[n]$ not in the image of $\alpha$ and let be the elements $j\in [m-1]$ with $\alpha (j)=\alpha (j+1)$. The monotone surjection $\sigma :={\sigma }^{j_1}\cdots {\sigma }^{j_s}:[m]\to [m-s]$ identifies exactly those adjacent pairs, and the monotone injection $\delta :={\delta }^{i_1}\cdots {\delta }^{i_r}:[n-r]\to [n]$ has exactly the complement of $\{i_1,\dots ,i_r\}$ as image. Since $\alpha$ is monotone, its image has $m-s+1$ elements, so $n-r=m-s$, and $\alpha$ factors through $\sigma$ by a monotone injection with the same image as $\alpha$, which is $\delta$. Uniqueness holds because the two index sets are determined by $\alpha$: the $i$’s are the values omitted, and the $j$’s the arguments at which $\alpha$ repeats. ◻

Proof. This is the Yoneda lemma (definition 141.58 and the development around it) for the presheaf $X$ on $\mathrm{\Delta }$. Concretely, a simplicial map $u:\mathrm{\Delta }[n]\to X$ is determined by $x:=u_n({\mathrm{id}}_{[n]})$, because naturality gives $u_m(\alpha )=u_m(\mathrm{\Delta }[n](\alpha )(\mathrm{id}))=X(\alpha )(x)$ for every $\alpha :[m]\to [n]$; and conversely $u_m(\alpha ):=X(\alpha )(x)$ defines a natural family, by functoriality of $X$. ◻

Proof. Given a simplicial map $u$, put $x_i:=u_{n-1}({\delta }^i)$; these are $(n-1)$-simplices because ${\delta }^i$ is an $(n-1)$-simplex of $\mathrm{\Delta }[n]$ lying in the horn for $i\ne k$. Naturality and equation 190.1 give $d_i{x}_j=u({\delta }^j\circ {\delta }^i)=u({\delta }^i\circ {\delta }^{j-1})=d_{j-1}{x}_i$ for , which is equation 190.6.

Conversely let such a family be given. Every simplex $\alpha$ of ${\mathrm{\Lambda }}_k^n$ factors as $\alpha ={\delta }^i\circ \beta$ for some $i\ne k$ outside the image of $\alpha$ and some monotone $\beta$, by lemma 190.4 applied to the corestriction of $\alpha$ to $[n]\setminus \{i\}$. Set $u(\alpha ):=X(\beta )(x_i)$. This is independent of the chosen $i$: if $\alpha$ also avoids $j\ne i$, write $\alpha ={\delta }^i\beta ={\delta }^j\gamma$, and then equation 190.1, equation 190.6 identify the two values, since both equal $X$ applied to the factorization of $\alpha$ through the codimension-two face ${\delta }^i{\delta }^{j-1}$. Naturality in $\alpha$ is immediate from functoriality of $X$, and the two constructions are mutually inverse because $u({\delta }^i)=X(\mathrm{id})(x_i)=x_i$.

The statement for $\partial \mathrm{\Delta }[n]$ is the same argument with the index $k$ not excluded. ◻

Proof. The case $n=2$, $k=1$. By lemma 190.12 and the calculation after it, a map ${\mathrm{\Lambda }}_1^2\to N(\mathcal{C})$ is a composable pair $c_0\stackrel{f_1}{\to }{c}_1\stackrel{f_2}{\to }{c}_2$. A filler is a $2$-simplex, that is a pair $(g_1,g_2)$ of composable arrows, with $d_2=f_1$ and $d_0=f_2$, hence $g_1=f_1$ and $g_2=f_2$; the remaining face $d_1$ is then forced to be $f_2\circ f_1$. So the filler exists and is unique.

The case $n=3$, $k=1$. A map ${\mathrm{\Lambda }}_1^3\to N(\mathcal{C})$ consists of the three $2$-simplices $x_0,x_2,x_3$ matching along their common edges. Reading off the edges, the data amount to arrows $f_1,f_2,f_3$ composable in that order together with the three composites recorded by $x_0,x_2,x_3$; a filler is a functor $[3]\to \mathcal{C}$, which is determined by $f_1,f_2,f_3$, and its face $d_1$ records $f_3\circ (f_2\circ f_1)$. Existence and uniqueness therefore hold, and the matching condition on the horn is exactly associativity.

General $n$ and . A functor $[n]\to \mathcal{C}$ is determined by the $n$ arrows $f_i:c_{i-1}\to c_i$, and each $f_i$ is an edge of the horn: the edge $⟨i-1,i⟩$ misses at least one element of $[n]$ other than $k$ as soon as $n\ge 2$. So a filler is unique if it exists. For existence, define the functor by the arrows read off the horn; its faces $d_j$ for $j\ne k$ agree with the given $x_j$ because both are the chains obtained by composing the same arrows, and $d_k$ is unconstrained. ◻

Proof. Suppose $\mathcal{C}$ is a groupoid. Inner horns fill by proposition 190.14. For ${\mathrm{\Lambda }}_0^2$ the data are two arrows $f_1:c_0\to c_1$ and $g:c_0\to c_2$ with a common source; the filler must be a composable pair with $d_2=f_1$ and $d_1=g$, so the second arrow is forced to be $g\circ f_1^{-1}$, which exists because $f_1$ is invertible. The case ${\mathrm{\Lambda }}_2^2$ is the same calculation with the arrow $f_2$ inverted instead. For $n\ge 3$ the argument of proposition 190.14 applies unchanged, because for $n\ge 3$ every edge $⟨i-1,i⟩$ lies in every horn, so the $n$ arrows are still determined and the missing face is still unconstrained.

Conversely suppose $N(\mathcal{C})$ is a Kan complex and let $f:c_0\to c_1$. The pair $(f,{\mathrm{id}}_{c_0})$, with $f$ on the edge $⟨01⟩$ and ${\mathrm{id}}_{c_0}$ on $⟨02⟩$, is a map ${\mathrm{\Lambda }}_0^2\to N(\mathcal{C})$ by lemma 190.12, since both edges start at $c_0$. A filler provides $g:c_1\to c_0$ with $g\circ f={\mathrm{id}}_{c_0}$. Applying the same argument to $g$ gives $h$ with $h\circ g={\mathrm{id}}_{c_1}$, and then $h=h\circ g\circ f=f$, so $g$ is a two-sided inverse of $f$. ◻

Proof. Write $|{\mathrm{\Lambda }}_k^n|\subseteq {\mathrm{\Delta }}^n$ for the union of the geometric faces ${\delta }_{\ast }^i({\mathrm{\Delta }}^{n-1})$ with $i\ne k$. By lemma 190.12 a map ${\mathrm{\Lambda }}_k^n\to \mathrm{Sing}(Y)$ is a family of continuous maps on those faces agreeing on their intersections, hence by lemma 187.11 — the faces are closed — a single continuous map $u:|{\mathrm{\Lambda }}_k^n|\to Y$. A filler is a continuous extension of $u$ to ${\mathrm{\Delta }}^n$. It suffices to produce a retraction $\rho :{\mathrm{\Delta }}^n\to |{\mathrm{\Lambda }}_k^n|$, for then $u\circ \rho$ is such an extension.

Let $m$ be the barycenter of ${\mathrm{\Delta }}^n$, let $b$ be the barycenter of the $k$-th face, and put $c:=2b-m$. Its coordinates satisfy $\sum _i{c}_i=2-1=1$, , and for $j\ne k$; so $c$ lies in the plane of ${\mathrm{\Delta }}^n$, beyond the $k$-th face.

For $x\in {\mathrm{\Delta }}^n$ consider $y(\lambda ):=c+\lambda (x-c)$, so that $y(1)=x$. Put $$\mu (x):=\underset{j\ne k}{max}max(0,\frac{c_j-x_j}{c_j}),\rho (x):=c+\frac{1}{\mu (x)}(x-c).$$ Here $\mu$ is continuous, being a maximum of finitely many continuous functions, and : if $x_j\ge c_j$ for every $j\ne k$ then , contradicting $\sum _i{x}_i=1$ and $x_k\ge 0$. Moreover $\mu (x)\le 1$, because each fraction is at most $1$ when $x_j\ge 0$. So $1/\mu (x)\ge 1$ and $\rho$ is continuous.

$\rho$ lands in the horn. Writing $\lambda :=1/\mu (x)$, the $j$-th coordinate of $\rho (x)$ is $c_j+\lambda (x_j-c_j)$, which vanishes exactly when $\lambda =c_j/(c_j-x_j)$, that is when the $j$-th entry realizes the maximum defining $\mu (x)$. Some $j\ne k$ does realize it, since ; for that $j$ the coordinate vanishes, so $\rho (x)$ lies in the $j$-th face, which is part of $|{\mathrm{\Lambda }}_k^n|$. All other coordinates of $\rho (x)$ are nonnegative, since each $j$-th entry of the maximum is at most $\mu (x)$, and the $k$-th coordinate is $c_k+\lambda (x_k-c_k)\ge c_k+(x_k-c_k)=x_k\ge 0$ because and $\lambda \ge 1$. So $\rho (x)\in {\mathrm{\Delta }}^n$.

$\rho$ fixes the horn. If $x_j=0$ for some $j\ne k$, then the $j$-th entry of the maximum is $1$, so $\mu (x)=1$ and $\rho (x)=x$. ◻

Proof. (i) Let $u:{\mathrm{\Lambda }}_k^n\to X{\times }_{Z}Y$ and $v:\mathrm{\Delta }[n]\to Y$ agree after $q$. Write $u=(u_X,u_Y)$ with $u_Y=v{|}_{{\mathrm{\Lambda }}_k^n}$. Then $u_X$ and $g\circ v$ form a lifting problem for $p$, since $p\circ u_X=g\circ u_Y$. A filler $w_X$ gives $(w_X,v)$, which lands in the pullback because $p\circ w_X=g\circ v$, restricts to $u$ on the horn, and satisfies $q\circ (w_X,v)=v$.

(ii) Given $p:X\to Y$, $p^{\prime }:Y\to Z$ and a problem $u:{\mathrm{\Lambda }}_k^n\to X$, $v:\mathrm{\Delta }[n]\to Z$ with $p^{\prime }pu=v|$, first fill for $p^{\prime }$ with the horn map $p\circ u$, obtaining $w^{\prime }:\mathrm{\Delta }[n]\to Y$; then fill for $p$ with $u$ and $w^{\prime }$.

(iii) A lifting problem for $p\times p^{\prime }$ is a pair of lifting problems, one for $p$ and one for $p^{\prime }$, because maps into a product are pairs of maps (computed dimensionwise). Fill each separately. The last claim is the case $Y=Y^{\prime }=\mathrm{\Delta }[0]$, whose product is $\mathrm{\Delta }[0]$.

(iv) Given $u:{\mathrm{\Lambda }}_k^n\to X$, fill the horn $p\circ u$ in $Y$ to get $v:\mathrm{\Delta }[n]\to Y$, then fill the resulting lifting problem for $p$. ◻

Proof. Reflexivity. The degenerate edge $s_{0}x$ has $d_0{s}_{0}x=x=d_1{s}_{0}x$ by equation 190.4, so $x\sim x$.

Transitivity. Let $ϵ$ be an edge from $x$ to $y$ and $\eta$ one from $y$ to $z$. Then $(x_0,x_2):=(\eta ,ϵ)$ is a map ${\mathrm{\Lambda }}_1^2\to X$, because $d_0{x}_2=y=d_1{x}_0$ is the matching condition of lemma 190.12. A filler is a $2$-simplex whose remaining face $d_1$ is an edge with $d_1{d}_1=d_1{d}_2=x$ and $d_0{d}_1=d_0{d}_0=z$, using equation 190.4. So $x\sim z$.

Symmetry. Let $ϵ$ be an edge from $x$ to $y$. Then $(x_1,x_2):=(s_{0}x,ϵ)$ is a map ${\mathrm{\Lambda }}_0^2\to X$: the matching condition for the indices reads $d_1{x}_2=d_1{x}_1$, and both sides are $x$. A filler has $d_0$ an edge with $d_1{d}_0=d_0{d}_2=y$ and $d_0{d}_0=d_0{d}_1=x$, so it is an edge from $y$ to $x$. ◻