Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 3.3.2.3. Let $n \geq 0$ be an integer. Then there is a unique homeomorphism of topological spaces

\[ f: | \operatorname{N}_{\bullet }( \operatorname{Chain}[n] ) | \rightarrow | \Delta ^{n} | \]

with the following properties:

$(1)$

For every nonempty subset $S \subseteq [n]$, the map $f$ carries $S$ (regarded as a vertex of $\operatorname{N}_{\bullet }( \operatorname{Chain}[n] )$) to the barycenter $b_{S} \in | \Delta ^{S} | \subseteq | \Delta ^{n} |$.

$(2)$

For every $m$-simplex $\sigma : \Delta ^{m} \rightarrow \operatorname{N}_{\bullet }( \operatorname{Chain}[n] )$, the composite map

\[ | \Delta ^{m} | \xrightarrow { | \sigma | } | \operatorname{N}_{\bullet }( \operatorname{Chain}[n] ) | \xrightarrow {f} | \Delta ^{n} | \]

is affine: that is, it extends to an $\operatorname{\mathbf{R}}$-linear map from $\operatorname{\mathbf{R}}^{m+1} \supseteq | \Delta ^{m} |$ to $\operatorname{\mathbf{R}}^{n+1} \supseteq | \Delta ^{n} |$.

Proof. Note that an affine map $| \Delta ^{m} | \rightarrow | \Delta ^{n} |$ is uniquely determined by its values on the vertices of the topological $m$-simplex $| \Delta ^{m} |$. From this observation, it is easy to deduce that there is a unique continuous function $f: | \operatorname{N}_{\bullet }( \operatorname{Chain}[n] ) | \rightarrow | \Delta ^{n} |$ which satisfies conditions $(1)$ and $(2)$ of Proposition 3.3.2.3. We will complete the proof by showing that $f$ is a homeomorphism. Since the domain and codomain of $f$ are compact Hausdorff spaces, it will suffice to show that $f$ is a bijection. Unwinding the definitions, this can be restated as follows:

$(\ast )$

For every point $(t_0, t_1, \ldots , t_ n) \in | \Delta ^{n} |$, there exists a unique chain $S_0 \subsetneq S_1 \subsetneq \cdots \subsetneq S_ m$ of subsets of $[n]$ and positive real numbers $(s_0, s_1, \ldots , s_ m)$ satisfying the identities

\[ \sum s_ i = 1 \quad \quad (t_0, t_1, \ldots , t_ n) = \sum s_ i b_{S_{i}}. \]

We will deduce $(\ast )$ from the following more general assertion:

$(\ast ')$

For every element $(t_0, t_1, \ldots , t_ n) \in \operatorname{\mathbf{R}}_{\geq 0}^{n+1}$, there exists a unique (possibly empty) chain $S_0 \subsetneq S_1 \subsetneq \cdots \subsetneq S_ m$ of subsets of $[n]$ and positive real numbers $(s_0, s_1, \ldots , s_ m)$ satisfying $(t_0, t_1, \ldots , t_ n) = \sum s_ i b_{S_{i}}.$

Note that, if $(t_0, t_1, \ldots , t_ n)$ and $(s_0, s_1, \ldots , s_ m)$ are as in $(\ast ')$, then we automatically have $\sum _{i=0}^{m} s_ i = \sum _{j=0}^{n} t_ j$. It follows that assertion $(\ast )$ is a special case of $(\ast ')$. To prove $(\ast ')$, let $K \subseteq [n]$ be the collection of those integers $j$ for which $t_ j \neq 0$. We proceed by induction on the cardinality of $k = |K|$. If $k=0$ is empty, there is nothing to prove. Otherwise, set $r = \min \{ k t_ i \} _{i \in K}$. We can then write

\[ (t_0, t_1, \ldots , t_ n) = r b_{K} + (t'_0, t'_1, \ldots , t'_ n) \]

for a unique sequence of nonnegative real numbers $(t'_0, \ldots , t'_ n )$. Applying our inductive hypothesis to the sequence $(t'_0, \ldots , t'_ n)$, we deduce that there is a unique chain of subsets $S_0 \subsetneq S_1 \subsetneq \cdots \subsetneq S_{m-1}$ of $[n]$ and positive real numbers $(s_0, s_1, \ldots , s_{m-1} )$ satisfying $(t'_0, t'_1, \ldots , t'_ n) = \sum s_ i b_{S_{i}}$. Note that each $S_{i}$ is contained in $K'$, and therefore properly contained in $K$. To complete the proof, we extend this sequence by setting $S_ m = K$ and $s_ m = r$. $\square$