Exercise 1.6.10 (If $|\Delta| = |\Omega|$, then $S_\Delta \simeq S_\Omega$)

Proof. Since $|\mathrm{\Delta }|=|\mathrm{\Omega }|$ , there is a bijection $𝜃:\mathrm{\Delta }\to \mathrm{\Omega }$. We define the map

so that the following diagram is commutative:

(Some problems with the Solverer Tex compilation with tikz: take $\mathrm{\Delta }$ for $D$ and $\mathrm{\Omega }$ for $O$)

(a)

Since ${𝜃}^{-1}:\mathrm{\Omega }\to \mathrm{\Delta }$, $\sigma :\mathrm{\Delta }\to \mathrm{\Delta }$, and $𝜃:\mathrm{\Delta }\to \mathrm{\Omega }$, then $𝜃\circ \sigma \circ {𝜃}^{-1}:\mathrm{\Omega }\to \mathrm{\Omega }$. Moreover, $𝜃$ and $\sigma$ are bijections, thus $𝜃\circ \sigma \circ {𝜃}^{-1}$ is a bijection, i.e. a permutation of $\mathrm{\Omega }$, so $\phi$ is well defined.

(b)

Consider the map $$\psi \{\begin{array}{ccc}\hfill S_{\mathrm{\Omega }}\hfill & \hfill \to \hfill & S_{\mathrm{\Delta }}\hfill \\ \hfill \tau \hfill & \hfill \mapsto \hfill & \psi (\tau )={𝜃}^{-1}\circ \tau \circ 𝜃.\hfill \end{array}$$

Then the following diagram is commutative

As in part (a), $\psi$ is well defined.

For all $\sigma \in S_{\mathrm{\Delta }}$,

$$\begin{array}{llll}\hfill (\psi \circ \phi )(\sigma )& =\psi (\phi (\sigma ))& \hfill & \\ \hfill & =\psi (𝜃\circ \sigma \circ {𝜃}^{-1})& \hfill & \\ \hfill & ={𝜃}^{-1}\circ (𝜃\circ \sigma \circ {𝜃}^{-1})\circ 𝜃& \hfill & \\ \hfill & =({𝜃}^{-1}\circ 𝜃)\circ \sigma \circ ({𝜃}^{-1}\circ 𝜃)& \hfill & \\ \hfill & =\sigma .& \hfill & \end{array}$$

Therefore $\psi \circ \phi =1_{S_{\mathrm{\Delta }}}$, and similarly $\phi \circ \psi =1_{S_{\mathrm{\Omega }}}$. This proves that $\phi$ is bijective, and ${\phi }^{-1}=\psi$.

(c)

We verify that $\phi$ is a homomorphism: for all $\sigma ,\tau \in S_{\mathrm{\Delta }}$, $$\begin{array}{llll}\hfill \phi (\sigma )\circ \phi (\tau )& =(𝜃\circ \sigma \circ {𝜃}^{-1})\circ (𝜃\circ \tau \circ {𝜃}^{-1})& \hfill & \\ \hfill & =𝜃\circ \sigma \circ ({𝜃}^{-1}\circ 𝜃)\circ \tau \circ {𝜃}^{-1})& \hfill & \\ \hfill & =𝜃\circ (\sigma \circ \tau )\circ {𝜃}^{-1}& \hfill & \\ \hfill & =\phi (\sigma \circ \tau ).& \hfill & \end{array}$$

By part (b) and (c), $\phi$ is an isomorphism, so $S_{\mathrm{\Delta }}\simeq S_{\mathrm{\Omega }}$. □