Question about the definition of left normal morphism of augmented algebras

Question

On the renowned "On the Structure of Hopf Algebras" by Milnor and Moore, there is a definition of "left normal morphism of augmented algebras." It says as follows. If $A$ and $B$ are augmented $R$-algebras, then $f:A\rightarrow B$ is called left normal if the natural map $\pi:B\rightarrow R\otimes_{A}B=B/(I(A)B)$ is a split epimorphism and $BI(A)\subset I(A)B$. But still I cannot understand why do we need $\pi$ to be split. Would someone explain me why do we need this?

Answer 1

I was looking at this too and I think I see why. So here's the setting :

$f \colon A \to B$ is a morphism of augmented $K$-algebras, and denote the augmentations by $\epsilon$. If $f$ is left normal, then there is a unique $K$-algebra structure on $C = K \otimes_A B$ making the natural projection $\pi \colon B \to K \otimes_A B \colon b \mapsto 1 \otimes b$ a morphism of augmented $K$-algebras.

Even without the splitting of $\pi$ you can define the product on $C$ just by $1 \otimes b \cdot 1 \otimes b' = 1 \otimes b b'$ and the unit just by composition $K \to B \to C \colon 1 \mapsto 1 \otimes 1$. This makes $C$ into an algebra.

Now for the augmentation, the map $C \to K \colon 1 \otimes b \mapsto 1 \cdot \epsilon(b)$ is not well-defined, since $K \times B \to K \colon (1,b) \mapsto 1 \epsilon(b)$ is not $A$-balanced. But if $\pi$ admits a splitting $j$, then you can define the counit on $C$ as $\epsilon \circ j$ making the diagram commute.

So that's interesting as it seems like the splitting is not needed for the algebra structure on the quotient, but for the augmented algebra structure, and eventually the Hopf algebra structure there.