In low dimensions:

• $|\Delta^0| = \{1\}$: a single point.
• $|\Delta^1| = \{(t_0, t_1) : t_0 + t_1 = 1, t_i \geq 0\} \cong [0, 1]$: a line segment.
• $|\Delta^2|$: a filled triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$.
• $|\Delta^3|$: a solid tetrahedron.

The coface map $|\delta^i|: |\Delta^{n-1}| \to |\Delta^n|$ embeds the $(n-1)$-simplex as the $i$-th face (the face opposite vertex $i$). The codegeneracy $|\sigma^j|: |\Delta^{n+1}| \to |\Delta^n|$ collapses the edge from vertex $j$ to vertex $j+1$.

The geometric realization $|\Delta[n]|$ is homeomorphic to $|\Delta^n|$ (the topological $n$-simplex). This follows from the Yoneda lemma and the coend formula:

$|\Delta[n]| = \int^{[m]} \operatorname{Hom}_\Delta([m], [n]) \times |\Delta^m| \cong |\Delta^n|.$

The last isomorphism is the "co-Yoneda lemma" (or density theorem for coends).

The geometric realization $|\partial\Delta[n]|$ is the boundary of the topological $n$-simplex, i.e., the union of all $(n-1)$-dimensional faces of $|\Delta^n|$. This is homeomorphic to the $(n-1)$-sphere $S^{n-1}$.

For $n = 2$: $|\partial\Delta[2]|$ is the boundary of a triangle, homeomorphic to $S^1$.

The simplicial set $S^1 = \Delta[1]/\partial\Delta[1]$ has one vertex and one non-degenerate edge (with both endpoints identified). Its geometric realization $|S^1|$ is obtained by taking the interval $[0,1]$ and identifying $0 \sim 1$, giving the circle $S^1$.

For a group $G$, the geometric realization $|N(BG)|$ is the classifying space $BG$.

For $G = \mathbb{Z}/n\mathbb{Z}$: $|N(B(\mathbb{Z}/n\mathbb{Z}))| \simeq L_n^\infty$, the infinite lens space, which has $\pi_1 = \mathbb{Z}/n\mathbb{Z}$ and is a $K(\mathbb{Z}/n\mathbb{Z}, 1)$.

For $G = S_3$ (symmetric group on 3 elements): the classifying space $BS_3$ has $\pi_1 = S_3$ and all higher homotopy groups trivial.

The product $\Delta[1] \times \Delta[1]$ is a simplicial set whose geometric realization is the square $[0,1]^2$. This square is triangulated: the non-degenerate $2$-simplices correspond to the two triangles in the standard triangulation of the square (upper-left and lower-right).

The Eilenberg--Zilber theorem provides a natural quasi-isomorphism between $C_*(X \times Y)$ and $C_*(X) \otimes C_*(Y)$ at the chain level.

Geometric realization preserves all colimits (since it is a left adjoint). In particular:

$|X \sqcup Y| = |X| \sqcup |Y|$

and more generally, if $X = \operatorname{colim}_i X_i$, then $|X| = \operatorname{colim}_i |X_i|$.

This means $|X|$ is built by gluing together the geometric realizations of the standard simplices appearing in $X$.

$\operatorname{Sing}(*)_n$ is a single element for each $n$ (the unique map $|\Delta^n| \to *$). So $\operatorname{Sing}(*) = \Delta[0]$, the terminal simplicial set.

$\operatorname{Sing}([0,1])$ is an "enormous" simplicial set: its $1$-simplices are all continuous paths $[0,1] \to [0,1]$, its $2$-simplices are all continuous maps from a triangle into $[0,1]$, etc. Despite being large, it is a Kan complex (every horn can be filled) and is weakly equivalent to a point (since $[0,1]$ is contractible).

$\operatorname{Sing}(S^1)$ is a Kan complex with $\pi_1 = \mathbb{Z}$ and $\pi_n = 0$ for $n \geq 2$. It has uncountably many $1$-simplices (all loops and paths in $S^1$), but its homotopy type is completely described by $\pi_1 = \mathbb{Z}$.

For any topological space $Y$, the counit $\varepsilon_Y: |\operatorname{Sing}(Y)| \to Y$ is a weak homotopy equivalence: it induces isomorphisms on all homotopy groups. This is a classical theorem in algebraic topology.

For CW complexes, this map is actually a homotopy equivalence (by Whitehead's theorem). For general spaces, $|\operatorname{Sing}(Y)|$ is a CW approximation of $Y$.

This means that from the perspective of homotopy theory, every topological space is "modeled" by a simplicial set.

The unit $\eta_X: X \to \operatorname{Sing}(|X|)$ is generally not an isomorphism (or even a weak equivalence for arbitrary $X$). However, when $X$ is a Kan complex, $\eta_X$ is a weak equivalence.

For example, if $X = \Delta[1]$, then $\operatorname{Sing}(|\Delta[1]|) = \operatorname{Sing}([0,1])$, which has uncountably many $1$-simplices, while $\Delta[1]$ has only three (two degenerate, one non-degenerate).