In the Milnor and Moore paper, "[On the structure of Hopf algebras](https://doi.org/10.2307/1970615)" 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  

$$A\otimes C \xrightarrow{\cong\otimes C } A\otimes K\otimes C \xrightarrow{A\otimes \eta\_N\otimes C } A\otimes N \otimes C \xrightarrow{\varphi\_N} N\otimes C. $$



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


