# Question about the definition of left normal morphism of augmented algebras

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?

$f \colon A \to B$ 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.
• Well.. I think the natural projection $\pi:B\rightarrow K\otimes_{A}B$ is $b\mapsto1\otimes b$, not $b\mapsto\epsilon(b)\otimes b$... Oct 21, 2013 at 5:10