TITLE: Is this almost-cosimplicial object familiar?
QUESTION [6 upvotes]: I have a sequence $X_0,X_1,\ldots$ of Abelian groups, along with face maps $d_0,\ldots, d_n: X_{n-1}\to X_n$ which satisfy the standard cosimplicial identities EXCEPT at the very bottom, where in $\mathrm{Hom}(X_0, X_2)$, I have
$$d_2d_0 \ne d_0d_1$$
However, I have one extra map $d_\infty:X_1\to X_2$ which satisfies the following relations in $\mathrm{Hom}(X_0,X_2)$
$$
d_\infty d_0 = d_2d_0
$$
$$
d_\infty d_1 = d_0d_1
$$
and the following relations in $\mathrm{Hom}(X_1,X_3)$
$$
d_0d_\infty = d_1d_\infty
$$
$$
d_2d_\infty = d_3d_\infty
$$
By hand, I can build a chain complex out of $X_\cdot$, adding a $d_\infty$ to the usual cosimplicial differential. The relations above imply that $d^2=0$.

Is this familiar to anyone?

At first I had hoped that I just had the "wrong" face maps and if I changed around my face maps $X_1\to X_2$ in a clever enough way that this would become a regular (semi)-cosimplicial Abelian group. But I don't see it. 
Another thought I had is that I have many other $d_\infty$ maps that I just haven't seen yet because I don't know the relations that they should satisfy and so am not looking in the right place.
I also noticed that $d_\infty$ satisfies the relations I'd expect from the map $[1]=\lbrace 0,1\rbrace$ to $[2]=\lbrace 0,1,2\rbrace$ which is the constant map to $1$. But this seems not to lead anywhere because of the failure of the face identities at the bottom; i.e., the maps $X_0\to X_1$ and $X_1\to X_2$ don't act like regular injections of ordered sets.

REPLY [6 votes]: I will also talk about simplicial things rather than cosimplicial things.  Furthermore, I will say "semi-trapezoidal" instead of "almost-simplicial". "Semi-" because there is no codegeneracy maps, and "trapezoidal" because an element $x$ of $X_2$ can be visualized as a trapezoid with four edges $d_2(x), d_\infty(x), d_0(x)$ and $d_1(x)$.
There is a way to make a semi-trapezoidal set into a semi-simplicial set. Given a semi-trapezoidal set $X$ define a semi-simplicial set $A$ setting
$A_0 = X_0, A_1 = X_1 + 2X_2, A_2 = 4X_2, A_3 = X_3 + X_2$ and $A_n = X_n$ for $n > 3.$
Roughly speaking, what we are doing is: We leave the $0$-simplices unchanged. For $1$-simplices we take those of $X$, and add to them the diagonals of the trapezoids (directed in a certain way). For the set of 2-simplices we take the set of triangles in which trapezoids are divided by their diagonals (there are four triangle for each trapezoid). Each trapezoid becomes a 3-simplex of $A$, as well as each original 3-simplex of $X$ becomes a 3-simplex of $A$. More concretely:  
Denote the two embeddings of $X_2 \rightarrow A_1$ by $i_a$ and $i_b$, and their respective images by $A_{a}$ and $A_{b}$. Denote the embeddings $X_2 \rightarrow A_2$ by $i_{aa}, i_{ab}, i_{ba}, i_{bb}$ and their respective images by $A_{aa}, A_{ab}, A_{ba}, A_{bb}$. Define the face maps $e$ of $A$ by:
On $X_1 \subset A_1$: $e_0 = d_0, e_1 = d_1$. 
On $A_{a}$: $e_0 = d_0d_0 = d_0d_1$, $e_1 = d_0d_\infty = d_0d_2$
On $A_{b}$: $e_0 = d_1d_\infty = d_1d_0$, $e_1 = d_1d_2 = d_1d_1$
On $A_{aa}$: $e_0 = i_a, e_1 = d_2, e_2 = d_2$, and similarly on $A_{ab}$, $A_{ba}$, $A_{bb}$.
On $X_2 \subset A_3$: $e_0 = i_{ab}, e_1 = i_{bb}, e_2 = i_{aa}, e_3 = i_{ba}$.
On $X_3 \subset A_3$: $e_0 = i_{ab}d_0, e_1 = i_{bb}d_1, e_2 = i_{aa}d_2, e_3 = i_{ba}d_3$.
On $A_n$: $e_i = d_i$.
This construction extends to a functor and works for simplicial objects in a category with coproduct.
If we are in an abelian category, we can define chain complexes $CX$ and $CA$. There is a morphism $CX \rightarrow CA$ whose component $CX_2 \rightarrow CA_2$ is defined by $i_{aa} + i_{ab}$, which should be a quasi-isomorphism if the intuition is not wrong.