Consider a Hopf algebra $H$ over $\mathbb C$ with $\varepsilon \colon H\to \mathbb C$ being its counit map. I am trying to figure out why the standard resolution of $\mathbb C$, seen as a right $H$-module via the counit map, is indeed a resolution. To my knowledge, this is well-known, from the book of Cartan and Eilenberg, where the standard resolution of an augmented algebra is considered. To be more precise, consider the chain complex
$$\cdots \longrightarrow H^{\otimes n+1} \longrightarrow H^{\otimes n}\longrightarrow\dots\longrightarrow H \otimes H\longrightarrow H \longrightarrow 0,$$ with differentials
\begin{align*} H^{\otimes n+1} &\xrightarrow{d_{n+1}} H^{\otimes n} \\ h_1\otimes \cdots \otimes h_{n+1} &\mapsto \varepsilon(h_1)h_2\otimes \cdots \otimes h_{n+1} + \sum_{i=1}^n(-1)^{i}h_1\otimes \cdots \otimes h_ih_{i+1} \otimes \cdots \otimes h_{n+1}. \end{align*}
By $\otimes$, I mean $\otimes_{\mathbb C}$. The augmented complex
$$\cdots \longrightarrow H^{\otimes n+1} \longrightarrow H^{\otimes n}\longrightarrow\dots\longrightarrow H \otimes H\longrightarrow H \longrightarrow \mathbb C \xrightarrow{\varepsilon} 0,$$ is a complex of right $H$-modules, where $H^{\otimes n}$ is a right $H$-module via the multiplication of an element of $H$ and the extreme right term in an elementary tensor of $H^{\otimes n}$.
My question is: how does one justifies that the augmented complex is exact? Is there an explicit chain homotopy to the identity chain map to zero? My guess (I am not an expert) was that the family $\{l_n: H^{\otimes n} \to H^{\otimes n+1}\}_{n\in \mathbb N}$ with $l_0(1_{\mathbb C})=1_H$ and $l_n(h_1\otimes h_2\otimes \cdots \otimes h_n)=1_H\otimes h_1\otimes h_2\otimes \cdots \otimes h_n$ should work, i.e. that $d_{n+2}l_{n+1}+l_nd_{n+1}=\text{id}_{H^{\otimes n+1}}$ for all $n\in \mathbb N$, but what I obtain in some lower dimension computation is as follows
$(d_2l_1+l_0d_1 )(h_1)=\epsilon(h_1) \neq h_1$
$(d_3l_2+l_1d_2)(h_1\otimes h_2)= 1_H\otimes \varepsilon(h_1)h_2 \neq h_1 \otimes h_2$, etc.
