Does geometric realization of a Δ-set collapse connected simplices to a single point?

Question

According to Wikipedia, the geometric realization of a Δ-set is defined as the following quotient space:

Each Δ-set has a corresponding geometric realization, defined as

$|S|=\left(\coprod _{{n=0}}^{{\infty }}S_{n}\times \Delta ^{n}\right)/_{{\sim }}$

where we declare that

$(\sigma ,d^{i}t)\sim (d_{i}\sigma ,t)\quad {\text{ for all }}\sigma \in S_{n},t\in \Delta ^{n-1}$.

Here, $\Delta ^{n}$ denotes the standard n-simplex, and

$d^{i}\colon \Delta ^{n-1}\rightarrow \Delta ^{n}$

is the inclusion of the i-th face. The geometric realization is a topological space with the quotient topology.

Interestingly, we equate here $(\sigma ,d^{i}t)\sim (d_{i}\sigma ,t)$ each simplex with its constituent simplices (faces), and then take the quotient space up to the equivalence.

What are the implications of this quotient? Let's have a loop of three edges $e_0$, $e_1$ and $e_2$, connected by three vertices $v_0$, $v_1$ and $v_2$. Let these correspond to standard 1-simplices $f_0$, $f_1$, $f_2$ and 0-simplices $w_0$, $w_1$ and $w_2$. Then we get equivalences:

$(e_0, f_0)\sim (v_0, w_0)$

$(e_0, f_0)\sim (v_1, w_1)$

$(e_1, f_1)\sim (v_1, w_1)$

$(e_1, f_1)\sim (v_2, w_2)$

$(e_2, f_2)\sim (v_2, w_2)$

$(e_2, f_2)\sim (v_0, w_0)$

But because of transitivity, this just leads to

$(e_0, f_0)\sim (e_1, f_1)\sim (e_2, f_2) \sim (v_0, w_0)\sim (v_1, w_1)\sim (v_2, w_2)$, and the space collapses to a trivial topology!

Indeed, it looks like with this definition, any connected component collapses to a single point.

Am I misunderstanding something here, or is that indeed what the geometric realization of a Δ-set is meant to do?

Answer 1

My understanding of what $\Delta ^{n}$ is, was wrong; it is a space of a geometrical n-simplex, so its elements are points, not whole simplices. Similarly, an inclusion map will include a n-simplex to some specific set of points inside n+1-simplex, not spit out just another whole simplex, like the face maps do.

Therefore, the actual equivalences are: (denoting $p \in Δ^0$ as the only point in the 0-simplex)

$(e_0, d^0p)\sim (v_0, p)$

$(e_0, d^1p)\sim (v_1, p)$

$(e_1, d^0p)\sim (v_1, p)$

$(e_1, d^1p)\sim (v_2, p)$

$(e_2, d^0p)\sim (v_2, p)$

$(e_2, d^1p)\sim (v_0, p)$

And indeed, that forms a beatiful triangle-shaped $S_1$.

Answer 2

Always remember in geometric realization of a simplicial set only the cells corresponding to nondegenerate simplices(simplices which do not come from lower degree simplices via degeneracy maps) remain intact. Everything else gets collapsed to lower dimensional cells.