# Todd Trimble Nerve of opposite category

This note has to do with showing that the “classifying space” of a category, i.e., the geometric realization of the nerve of a category, is homeomorphic to that of the opposite category.

Let $i:\Delta \to \mathrm{Cat}$ be the full inclusion of finite nonempty ordinals into categories. The nerve functor $N$ is the “restricted Yoneda embedding”

$\mathrm{Cat}\stackrel{y}{\to }{\mathrm{Set}}^{{\mathrm{Cat}}^{\mathrm{op}}}\stackrel{{\mathrm{Set}}^{{i}^{\mathrm{op}}}}{\to }{\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}.$Cat \stackrel{y}{\to} Set^{Cat^{op}} \stackrel{Set^{i^{op}}}{\to} Set^{\Delta^{op}}.

On $\mathrm{Cat}$ there is an involution given by taking opposite categories, and similarly there is an involution on $\Delta$. Denote either involution by $\sigma$. There is an evident isomorphism

$\begin{array}{ccc}\mathrm{Cat}& \stackrel{N}{\to }& {\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}\\ {}^{\sigma }↓& \cong & {↓}^{{\mathrm{Set}}^{{\sigma }^{\mathrm{op}}}}\\ \mathrm{Cat}& \underset{N}{\to }& {\mathrm{Set}}^{{\Delta }^{\mathrm{op}}},\end{array}$\array{ Cat & \stackrel{N}{\to} & Set^{\Delta^{op}} \\ ^\mathllap{\sigma} \downarrow & \cong & \downarrow^\mathrlap{Set^{\sigma^{op}}} \\ Cat & \underset{N}{\to} & Set^{\Delta^{op}}, }

i.e., the nerve functor commutes with these involutions. The vertical arrow on the right takes a simplicial set $X$ to $X\circ \sigma$; we may call this the opposite simplicial set.

Geometric realization is a functor

$R:{\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}\to \mathrm{Top}$R: Set^{\Delta^{op}} \to Top

where $\mathrm{Top}$ is any convenient category of spaces, such as $\mathrm{CG}\mathrm{Haus}$, and $R$ is defined by a coend or tensor product

$R\left(X\right)={\int }^{n}X\left(n\right)\cdot \mathrm{Int}\left(\left[n\right],I\right).$R(X) = \int^n X(n) \cdot Int([n], I).

In this formula, $\left[n\right]\in \mathrm{Ob}\left({\Delta }^{\mathrm{op}}\right)$ is regarded as an interval consisting of $n+1$ totally ordered elements, and $I$ as a topological interval. The hom-object $\mathrm{Int}\left(\left[n\right],I\right)$ is regarded as living in $\mathrm{Top}$.

If $\sigma I$ is the interval $I$ with reverse ordering, then there is a homeomorphism

$\mathrm{Int}\left(\left[n\right],I\right)\cong \mathrm{Int}\left(\sigma \left[n\right],\sigma I\right)$Int([n], I) \cong Int(\sigma [n], \sigma I)

and there is also an interval homeomorphism $I\to \sigma I$ sending $t$ to $1-t$. Thus we have

$\begin{array}{ccc}{\int }^{n}X\left(n\right)\cdot \mathrm{Int}\left(\left[n\right],I\right)& \cong & {\int }^{n}X\left(n\right)\cdot \mathrm{Int}\left(\sigma \left[n\right],\sigma I\right)\\ & \cong & {\int }^{n}X\left(n\right)\cdot \mathrm{Int}\left(\sigma \left[n\right],I\right)\\ & \cong & {\int }^{n}X\sigma \left(n\right)\cdot \mathrm{Int}\left(\left[n\right],I\right)\end{array}$\array{ \int^n X(n) \cdot Int([n], I) & \cong & \int^n X(n) \cdot Int(\sigma [n], \sigma I) \\ & \cong & \int^n X(n) \cdot Int(\sigma [n], I) \\ & \cong & \int^n X \sigma (n) \cdot Int([n], I) }

Hence the geometric realization of a simplicial set $X$ is homeomorphic to the geometric realization of the opposite simplicial set. It follows that the classifying space of a category is homeomorphic to the classifying space of its opposite.

Revised on November 24, 2011 01:48:56 by Todd Trimble