Generic example of a simplicial set

Let $X$ be a topological space. Define

$$\sS_n(X):=\{f\:\Dl\to X,\mbox{ continuous}\}$$

We claim that $\sS_{\bullet}(X)$ is a simplicial set with the following face and degeneracy maps:
$$\begin{align*} d_i\: \sS_n(X) \to \sS_{n-1}(X) &,\quad d_i(f):=f\circ \dl_i \\ s_j\: \sS_n(X) \to \sS_{n+1}(X) &,\quad s_j(f):=f\circ \sg_j \\ \end{align*}$$
It is called singular functor. It goes from the category of topological spaces to the category of simplicial sets.

$$\sS_{\bullet}(-)\: \Top \to \SSets$$

Recall the functor of geometric realization of a simplicial set,

$$K_{\bullet} \mapsto \vert K_{\bullet}\vert,\quad \vert-\vert\:\SSets \to \Top$$

In the example () $R$-modules can be replaced by functors. Left modules correspond to covariant functors, and right modules correspond to contravariant functors. Then the geometric realization functor can be seen as a tensor product over the simplicial category

$$\vert K_{\bullet}\vert=K_{\bullet}\ox_{\Dl}\Dl^{\bullet}$$

In an analogous way we can present the singular functor as

$$\sS_{\bullet}(X)=\Hom_{\Top}(\Dl^{\bullet}, X)$$

Hence we can derive adjointness

$$\Hom_{\Top}(K_{\bullet}\ox_{\Dl}\Dl^{\bullet}, X)\isom \Hom_{\Dl}(K_{\bullet}, \Hom_{\Top}(\Dl^{\bullet}, X))$$

Now the question arises: how to compare $X$ and $\vert\sS_{\bullet}(X)\vert$? Take $\id\in\Hom_{\SSets}(\sS_{\bullet}(X), \sS_{\bullet}(X))$. This identity goes to a map

$$\eps\: \vert\sS_{\bullet}(X)\vert\to X$$

which is called a unit. Also $\id\in\Hom_{\Top}(\vert K_{\bullet}\vert, \vert K_{\bullet}\vert)$ goes to a map

$$\eta\: K_{\bullet}\to \sS_{\bullet}(\vert K_{\bullet}\vert)$$

which is called a counit. If $X$ is a CW-complex, then this map is a homotopy equivalence.

Now we will prove the following theorem.

Before the proof, we will give some necessary propositions.



Let $C_{\bullet}$ be the cyclic set, whose geometric realization is the circle. Naive way to define an $S^1$-action would be to use

$$C_{\bullet}\times X_{\bullet}\to X_{\bullet}$$

$$(g, x)\mapsto g_*(x)$$

But it does not work, since it gives a trivial action of $S^1$ for $X_{\bullet}=C_{\bullet}$.

There is a forgetful functor from the category of cyclic sets to the category of simplicial sets.

$$G\:\CSets\to\SSets$$

We will define its left adjoint

$$F\: \SSets\to \CSets$$

If $Y_{\bullet}$ is a simplicial set, then put

$$F(Y_{\bullet})_n:= C_n\times Y_n,\quad C_n=\bZ/(n+1)\bZ$$

If $f$ is a morphism in $\Dl^{op}$, then we define

$$f_*(g, y):= (f_*(g), (g^*(f))_*(y))$$

$$\xymatrix{ [n] \ar[r]^{f} \ar[d]_{f_*(g)} & [m] \ar[d]^{g} \\ [n] \ar[r]_{g^*(f)} & [m] }$$

If $h$ is a morphism in $C_m$, then we define

$$h^*(g, y):=(h(g), y)$$

Observe that the composite

$$\vert F(X_{\bullet})\vert\to \vert C_{\bullet}\vert\times \v... ...rt\xrightarrow{\isom} \vert C_{\bullet}\times X_{\bullet}\vert$$

is not the geometric realization of a simplicial map.