Proof

The algorithmic formulae used when proving that any simplicial group is a Kan complex (cf., entry on simplicial group) give a filler defined as a product of degenerate copies of the faces of the given box, so is in $D_n$.

Proof

To see this, we note that as every degenerate element is this, $D \subseteq T$. Conversely if $t \in T_n$, then it fills the box made up of $( - , d_1t, \ldots, d_n t )$. This, in turn, has a filler, $d$, in $D$, but, as this filler is also thin, it must be that $t = d$, since thin fillers are uniquely determined.

Proof

One way around, this is nearly trivial. If $(G,D)$ is a group $T$-complex and $x\in NG_n$, then $x$ fills a box $(-, 1, \ldots, 1)$, so if $x\in NG_n\cap D_n$, $x$ must itself be the thin filler, however 1 is also a thin filler for this box, so $x = 1$ as required.

Conversely if $NG\cap D = \{1\}$, then we must check the other two axioms as the first is trivial. As any box has a standard filler in $D$, we only have to check uniqueness, but if $x$ and $y$ are in $D_n$, and both fill the same box (with the $k^{th}$ face missing) then $z = xy^{-1}$ fills a box with 1s on all faces (and the $k^{th}$ face missing).

If $k = 0$, then as $z \in NG_n\cap D_n$, we have $z = 1$ and $x$ and $y$ are equal. If $k \gt 0$, assume that if $\ell \lt k$ and $z \in D_n \cap \bigcap_{i\neq \ell} Ker\, d_i$ then $z = 1$, (i.e, that we have uniqueness up to at least the $(k-1)^{st}$ case). Consider $w = zs_{k-1}d_k z^{-1}$. This is still in $D_n$ and $d_i w = 1$ unless $i = k-1$, hence by assumption $w = 1$. Of course, this implies that $z = s_{k-1}d_kz$, but then $d_{k-1} z = d_k z$. We know that $d_{k-1} z = 1$, so $d_k z = 1$ and $z = 1$, i.e., $x = y$ and we have uniqueness at the next stage.

To verify the third axiom, assume that $x \in D_{n + 1}$ and each $d_i x \in D_n$ for $i \neq k$, then we can assume that $k = 0$, since otherwise we can skew the situation around as before to get that to be true, verify it in that case and ‘skew’ it back again later.

Suppose therefore that $d_i x \in D_n$ for all $0 \lt i \lt n$. As $x$ must be the degenerate filler given by the standard method, we can calculate $x$ as follows:

let $w_n = s_{n-1}d_n x$, $w_i = w_{i+1}(s_{i-1}d_i w_{i+1})^{-1}s_{i-1}y_i$ for $i = 1$, then $x = w_1$. We can therefore check that $d_0x \in D_n$ as required.