Milnor and Moore paper "On the structure of Hopf algebra" proposition 1.7 , filtration and grading

Question

In the Milnor and Moore paper, "On the structure of Hopf algebras" proposition 1.7 said the following:

1. $A$ a connected $K$-algebra.
2. $N$ a left $A$ module that is connected as a $K$-graded module i.e. there is an isomorphism $\eta_N:K\to N_0$ which results in an augmentation $\varepsilon_N:N\to K$.
3. $C=K\otimes_A N$, here $K$ is considered as $A$-rightmodule by the action $K\otimes A \xrightarrow{\cong} A \xrightarrow{\varepsilon_A} K$.
4. $\Delta :N \to N\otimes C$ a morphism of left $A$ modules.
5. $\pi:N\to C$ the canonical epimorphism. Let $f:C\to N$ such that $\pi f = \operatorname{id}_C$. In this paper they used $C$ the object letter to indicate the identity morphism.
6. $(\varepsilon_N\otimes C)\circ \Delta =\pi$.
7. $(N\otimes \varepsilon_C)\circ \Delta = N$.
8. $A\otimes C \xrightarrow{i\otimes C} N\otimes C $ is a monomorphism.

If $\tilde{f}$ is the composition $$A\otimes C \xrightarrow{A\otimes f} A \otimes N \xrightarrow{\varphi_N} N$$ then $\tilde f$ is an isomorphism.

The map $i$ is defined as the composition $$A\xrightarrow{\cong} A\otimes K \xrightarrow{A\otimes \eta_N} A\otimes N \xrightarrow{\varphi_N} N$$ and $\varphi_N$ is the left action of $A$ on $N$.

My first question: in the process to show that $\tilde{f}$ is a monomorphism, they showed $\Delta \tilde{f}$ is a monomorphism.

They define a filtration on $$F_p(A\otimes C) = \sum_{q\leq p} A\otimes C_q \text{ and } F_p(N\otimes C) = \sum_{q\leq p} N\otimes C_q$$ Then they said: "let $E^0 (A\otimes C)$ be the associated bigraded module." Is this the grading associated with the filtration i.e. $\operatorname{gr}_p(A\otimes C) = F_p/F_{p-1}$? Or it is just the grading on $A$ and $C$ as $K$-modules? I guess it is the second since they said $E^0_{p,q} (A\otimes C) = A_p \otimes C_q$.

After that they said we identify $E^0(A\otimes C)$ with $A\otimes C$. Then why did they introduce the $E^0$?

Second question: How come the two following morphisms are the same: $$A\otimes C \xrightarrow{A\otimes f} A\otimes N \xrightarrow{\varphi_N} N\to_{\Delta} N\otimes C$$ and $i\otimes C$ which is
$$A\otimes C \xrightarrow{\cong\otimes C } A\otimes K\otimes C \xrightarrow{A\otimes \eta_N\otimes C } A\otimes N \otimes C \xrightarrow{\varphi_N} N\otimes C. $$

Answer 1

Although $E^0(A\otimes C)$ can be identified with $A\otimes C$ as objects, this is not compatible with morphisms. If $g\colon A\otimes C\to A\otimes C$ is a map of graded groups, then the map $g\colon A_i\otimes C_j\to(A\otimes C)_{i+j}$ can be expressed as a sum of morphisms $g_k\colon A_i\otimes C_j\to A_{i+k}\otimes C_{i-k}$. If $g$ preserves the filtration then $g_k=0$ for all $k<0$, and $E^0(g)=g_0$, whereas $g=\sum_{k\geq 0}g_k$.

In the case of interest, we have $g=\Delta\circ\widetilde{f}$. For $a\in A_i$ and $c\in C_j$ we have $g(a\otimes c)=\Delta(a\,f(c))$, but $\Delta$ is a map of $A$-modules by condition 4, so $g(a)=a\,\Delta(f(c))$. Here $\Delta(f(c))$ can be written as $\sum_{k=0}^jx_k$ with $x_k\in N_k\otimes C_{j-k}$ so $g_k(a\otimes c)=ax_k$ and $E^0(g)(a\otimes c)=g_0(a\otimes c)=ax_0$. On the other hand, $\epsilon$ is zero on $N_k$ for $k>0$ and $N_0=K$ so $x_0$ is essentially the same as $(\epsilon\otimes 1)(\Delta(f(c)))$, which is $\pi(f(c))=c$ by condition 6. Here we have implicitly identified $C$ with $K\otimes C\leq A\otimes C$; if we unwrap that we get $x_0=1\otimes c$ and $ax_0=a\otimes c$ in $A\otimes C$. This means that $E^0(\Delta\circ\widetilde{f})$ is the identity.