Calculating the homology of $\mathfrak{sl}_2$ Let $\mathcal{A} = \mathcal{U}(\mathfrak{sl}_2)$ (that is, the universal enveloping algebra of $\mathfrak{sl}_2$), $Z = \{x,h,y\}$ and $Z_{\mathbb{F}}$ be the $\mathbb{F}$-subspace of $\mathbb{F}\langle Z \rangle$ spanned by $Z$. Moreover $\mathcal{A}$ acts (from the right side) on $Z_{\mathbb{F}} \otimes_{\mathbb F} \mathcal{A}$ as follows: $(x \otimes y) \cdot \alpha = x \otimes y\alpha$. Hence $\mathcal{A}$ and $Z_{\mathbb{F}} \otimes_{\mathbb F} \mathcal{A}$ are right $\mathcal{A}$-modules. Finally, see $\mathbb{F}$ as a left $\mathcal{A}$-module with the trivial action (only the field $\mathbb{F}$ embedded in $\mathcal{A}$ effectively acts on $\mathbb{F}$).
Now let $\delta_0: Z_{\mathbb{F}} \otimes_{\mathbb F} \mathcal{A} \to \mathcal{A}$ be the $\mathcal{A}$-module homomorphism such that $\delta_0(w \otimes 1) = 1 \otimes w$ for any $w \in Z$ (here we are identifying $\mathcal{A} \cong \mathbb{F} \otimes_{\mathbb{F}} \mathcal{A}$).
Consider the following (module? abelian group?) homomorphism:
\begin{equation}
\mathcal{A} \otimes_{\mathcal{A}} \mathbb{F} \xleftarrow{\delta_0 \otimes 1} (Z_{\mathbb{F}} \otimes_{\mathbb F} \mathcal{A}) \otimes_{\mathcal{A}} \mathbb{F}
\end{equation}

My question is: what is $\ker(\delta_0 \otimes 1)$?

Here's my attempt: an arbitrary element of $(Z_{\mathbb{F}} \otimes_{\mathbb F} \mathcal{A}) \otimes_{\mathcal{A}} \mathbb{F}$ is of the form $\sum_i (x \otimes \alpha_i) \otimes \beta_i + \sum_j (h \otimes \alpha_j) \otimes \beta_j + \sum_i (y \otimes \alpha_i) \otimes \beta_i$. Applying $\delta_0 \otimes 1$ on that generic element gives us $ (1 \otimes x(\sum_i\alpha_i)) \otimes \beta_i + (1 \otimes  h(\sum_j\alpha_j)) \otimes \beta_j +  (1 \otimes y(\sum_k\alpha_k)) \otimes \beta_i$.
And that's it. I'm aware that $\ker(\delta_0) \otimes \mathbb{F} \subseteq \ker(\delta_0 \otimes 1)$ and also that certain choices of $\alpha_i,\alpha_j,\alpha_k,\beta_i,\beta_j,\beta_k$ will give us elements in the kernel (for example, the element $(1 \otimes xy) \otimes 1 + (1 \otimes  h(-1)) \otimes 1 +  (1 \otimes y(-x)) \otimes 1$ is an element in $\ker(\delta_0 \otimes 1)$). However I fail to express all elements in this set.
For context: I'm trying to calculate the homology of the Lie algebra $\mathcal{U}(\mathfrak{sl}_2)$ (which consists of the quotients $\ker(\delta_i \otimes 1)/im(\delta_{i+1} \otimes 1)$) using Anick's resolution. Calculating the images of $\delta_i \otimes 1$ is straightforward but the kernels are not.
 A: We have an isomorphism of vector spaces
$$
  \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}
  \to
  \mathbb{F} \,,
  \quad
  w \otimes \lambda
  \mapsto
  w \cdot \lambda \,,
$$
where the multiplication comes from the left $\mathcal{A}$-module structure on $\mathbb{F}$.
We have similarly an isomorphism
$$
  Z_{\mathbb{F}} \otimes_{\mathbb{F}} \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}
  \to
  Z_{\mathbb{F}} \otimes_{\mathbb{F}} \mathbb{F} \,,
  \quad
  w \otimes \alpha \otimes \lambda
  \mapsto
  w \otimes (\alpha \cdot \lambda) \,.
$$
Under these isomorphisms the linear map $\delta_0 \otimes 1$ corresponds to a linear map
$$
  \partial
  \colon
  Z_{\mathbb{F}} \otimes_{\mathbb{F}} \mathbb{F}
  \to
  \mathbb{F} \,.
$$
To better understand this linear map $\partial$ let $w \otimes \lambda$ be a simple tensor in $Z_{\mathbb{F}} \otimes_{\mathbb{F}} \mathbb{F}$.
The corresponding element in $\mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}$ is given by $w \otimes 1 \otimes \lambda$.
This element is mapped by $\delta_0 \otimes 1$ to the element $w \otimes \lambda$ of $\mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}$.
This last tensor product is taken over $\mathcal{A}$, whence we have
$$
  w \otimes \lambda
  =
  (1 \cdot w) \otimes \lambda
  =
  1 \otimes (w \cdot \lambda) \,.
$$
This element of $\mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}$ corresponds under the above isomorphism to the element $w \cdot \lambda$ of $\mathbb{F}$.
The linear map $\partial$ is thus given by
$$
  \partial
  \colon
  Z_{\mathbb{F}} \otimes \mathbb{F}
  \to
  \mathbb{F} \,,
  \quad
  w \otimes \lambda
  \mapsto
  w \cdot \lambda \,.
$$
But the action of $Z_{\mathbb{F}}$ os $\mathbb{F}$ is trivial, so
$$
  \partial
  =
  0
$$
and thus $\delta_0 \otimes 1 = 0$.

I want to point out that the above calculation is the same that we get when we work with the Koszul resolution of $\mathfrak{sl}_2$.
(I don’t know Anick’s resolution, so I can’t say how these two resolutions compare to each other.)
Indeed, let us abbreviate $\mathfrak{sl}_2$ as $\mathfrak{g}$.
The set $Z$ is a basis of $\mathfrak{g}$, and the vector space $Z_{\mathbb{F}}$ has $Z$ is a basis.
In other words, $Z_{\mathbb{F}}$ is basically just $\mathfrak{g}$ again.
If we use $\mathfrak{g}$ instead of $Z_{\mathbb{F}}$ then the map $\delta_0$ is given by
$$
  \delta_0
  \colon
  \mathfrak{g} \otimes_{\mathbb{F}} \mathcal{A}
  \to
  \mathcal{A} \,,
  \quad
  w \otimes \alpha
  \mapsto
  w \alpha \,.
$$
This map is part of the Koszul resolution
$$
  \dotsb
  \to
  \bigwedge^3(\mathfrak{g}) \otimes_{\mathbb{F}} \mathcal{A}
  \xrightarrow{\;\delta_2\;}
  \bigwedge^2(\mathfrak{g}) \otimes_{\mathbb{F}} \mathcal{A}
  \xrightarrow{\;\delta_1\;}
  \mathfrak{g} \otimes_{\mathbb{F}} \mathcal{A}
  \xrightarrow{\;\delta_0\;}
  \mathcal{A}
  \xrightarrow{\;\varepsilon\;}
  \mathbb{F}
  \to
  0 \,.
$$
If we were to use the Koszul resolution to compute the Lie algebra homology $\operatorname{H}^{\mathrm{Lie}}_*(\mathfrak{\mathfrak{g}}, \mathbb{F})$ then we would apply the functor $(-) \otimes_{\mathcal{A}} \mathbb{F}$ to this resolution and get the chain complex
$$
  \dotsb
  \to
  \bigwedge^2(\mathfrak{g}) \otimes_{\mathbb{F}} \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}
  \xrightarrow{\;\delta_1 \otimes 1\;}
  \mathfrak{g} \otimes_{\mathbb{F}} \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}
  \xrightarrow{\;\delta_0 \otimes 1\;}
  \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F}
  \to
  0 \,.
$$
By using an isomorphism of vector spaces $\bigwedge^n(\mathfrak{g}) \otimes_{\mathbb{F}} \mathcal{A} \otimes_{\mathcal{A}} \mathbb{F} \cong \bigwedge^n(\mathfrak{g}) \otimes_{\mathbb{F}} \mathbb{F}$ as above we would then then get a chain complex of the form
$$
  \dotsb
  \to
  \bigwedge^2(\mathfrak{g}) \otimes_{\mathbb{F}} \mathbb{F}
  \xrightarrow{\;\partial_1\;}
  \mathfrak{g} \otimes_{\mathbb{F}} \mathbb{F}
  \xrightarrow{\;\partial_0\;}
  \mathbb{F}
  \to
  0 \,.
$$
The linear map $\partial_0$ is then the same as the linear map $\partial$ as above.
