Calculating a specific example of a simplicial resolution of an algebra

Question

In the Stacks Project Tag 09D4, there's an explicit description of a simplicial resolution \begin{align*} P_{2}\to P_{1}\to P_{0} \end{align*} of an $A$-algebra $B$. I am trying to follow this construction step by step in an example $B=A[\varepsilon]/\varepsilon^{2}$. Let me describe what I have done so far, referring to the Stacks Project in turn.

First, one chooses a surjection $P_{0}=A[u_{i}\mid i\in I]\twoheadrightarrow B$. For our example here, $P_{0}=A[u]$ with surjection $u\mapsto\varepsilon$.

Second, one chooses generators $(f_{t}\mid t\in T)=\ker(P_{0}\twoheadrightarrow B)$. Here, $(f_{t})=(u^{2})$.

Third, one chooses $P_{1}=A[u_{i},x_{t}]$ where \begin{align*} d_{0}(u_{i})&=u_{i}&d_{1}(u_{i})&=u_{i}\\ d_{0}(x_{t})&=0&d_{1}(x_{t})&=f_{t}. \end{align*} Here, $P_{1}=A[u,x]$ and the face maps are \begin{align*} d_{0}(u)&=u&d_{1}(u)&=u\\ d_{0}(x)&=0&d_{1}(x)&=u^{2}. \end{align*}

Fourth, one considers the coskeleton \begin{align*} \mathrm{cosk}_{1}(P_{\bullet})_{2}&=\{(g_{0},g_{1},g_{2})\in P_{1}^{3}\mid d_{0}(g_{0})=d_{0}(g_{1}),d_{1}(g_{0})=d_{0}(g_{2}),d_{1}(g_{1})=d_{1}(g_{2})\} \end{align*} which has elements $y_{t}=(x_{t},x_{t},f_{t})$ and $z_{t}=(0,x_{t},x_{t})$, and in fact every element is $H+(0,0,g)$ for $H\in\mathrm{im}(A[u_{i},y_{t},z_{t}]\to\mathrm{cosk}_{1}(P_{\bullet})_{2})$ and $g=x_{t}x_{t'}-f_{t}x_{t'}$ or $g=\sum r_{t}x_{t}$ for $r_{t}\in P_{0}$ such that $\sum r_{t}f_{t}=0$.

Fifth, $P_{2}=A[u_{i},y_{t},z_{t},v_{r},w_{t,t'}]$ where $r=(r_{t})\in\ker(\bigoplus_{t}P_{0}\to P_{0})$ defined by $(r_{t})\mapsto \sum r_{t}f_{t}$. For us this map is $r\mapsto ru^{2}$ whose kernel is $(0)$ so $v_{r}=v_{0}=v$ and overall we have \begin{align*} P_{2}=A[u,y,z,v,w]. \end{align*}

Sixth, we have a map $P_{2}\to\mathrm{cosk}_{1}(P_{\bullet})_{2}$ defined by \begin{align*} y_{t}&\mapsto(x_{t},x_{t},f_{t})\\ z_{t}&\mapsto(0,x_{t},x_{t})\\ v_{r}&\mapsto(0,0,\sum r_{t}f_{t})\\ w_{t,t'}&\mapsto(0,0,x_{t}x_{t'}-f_{t}x_{t'}) \end{align*} which for us is \begin{align*} y&\mapsto(x,x,u^{2})\\ z&\mapsto(0,x,x)\\ v&\mapsto(0,0,0)\\ w&\mapsto(0,0,x^{2}-u^{2}x). \end{align*}

Question 1: Is everything I have so far correct? I don't discount I could be making a silly error somewhere.

Seventh, the Stacks Project claims that the map $P_{2}\to\mathrm{cosk}_{1}(P_{\bullet})_{2}$ determines the face maps $d_{0},d_{1},d_{2}\colon P_{2}\to P_{1}$.

Question 2: How does this work for us? In particular, for $A[u,y,z,v,w]\to A[u,x]$, what are $d_{0}$, $d_{1}$, and $d_{2}$ on the generators $u,y,z,v,w$?

Answer 1

I received help from Brian Shin on discord here (NB: that link may require you to join the AT discord server, but I am reproducing the conversation here for posterity).

Brian said:

To continue with your computation, I think the maps $d_i : P_2 \to P_1$ come from composing the map $P_2 \to \operatorname{cosk}_1(P_\bullet)_2$ with the face maps $d_i : \operatorname{cosk}_1(P_\bullet)_2 \to \operatorname{cosk}_1(P_\bullet)_1 = P_1$. Luckily, these face maps are given at Stacks 0188: $d_i(\alpha_0, \alpha_1, \alpha_2) = \alpha_i$.

Then I responded:

Thanks! So just to be explicit: $P_{2}\to P_{1}$ has \begin{align} d_{0}(u+y+z+v+w)&=u+x+0+0+0\\ d_{1}(u+y+z+v+w)&=u+x+x+0+0\\ d_{2}(u+y+z+v+w)&=u+u^{2}+x+0+x^{2}-u^{2}x \end{align} Does that look right?

Brian:

As far as I can tell, yes that looks right