We check that \(f * g\) preserves multiplication, the identity \(1_A\), and is \(R\)-linear (the last is automatic from the definition \(f * g = m_B \circ (f \otimes g) \circ \Delta\) being a composition of linear maps).

Identity. By the unit axiom of the bialgebra, \(\Delta(1_A) = 1_A \otimes 1_A\). Hence \((f * g)(1_A) = m_B(f(1_A) \otimes g(1_A)) = 1_B \cdot 1_B = 1_B\) since \(f, g\) are algebra homs.

Multiplication. For \(a, b \in A\), the bialgebra axiom \(\Delta(ab) = \Delta(a) \cdot \Delta(b)\) (i.e., \(\Delta\) is an algebra homomorphism with respect to the standard algebra structure on \(A \otimes A\)) gives

\[ \Delta(ab) \;=\; \sum_{(a),(b)} (a_{(1)} b_{(1)}) \otimes (a_{(2)} b_{(2)}). \]

Applying \((f * g)\):

\[ (f * g)(ab) \;=\; \sum_{(a),(b)} f(a_{(1)} b_{(1)}) \cdot g(a_{(2)} b_{(2)}) \;=\; \sum_{(a),(b)} f(a_{(1)})\,f(b_{(1)})\,g(a_{(2)})\,g(b_{(2)}), \]

using that \(f, g\) are algebra homs. Now use the commutativity of \(B\) to rearrange: \(f(b_{(1)})\,g(a_{(2)}) = g(a_{(2)})\,f(b_{(1)})\). This regroups the sum into

\[ \biggl(\sum_{(a)} f(a_{(1)}) g(a_{(2)})\biggr) \cdot \biggl(\sum_{(b)} f(b_{(1)}) g(b_{(2)})\biggr) \;=\; (f * g)(a) \cdot (f * g)(b). \]

Hence \(f * g\) preserves multiplication and is therefore an algebra homomorphism.
\(\square\)

Remark. Without the commutativity of \(B\), the rearrangement step fails, and \(f * g\) is only an \(R\)-linear map in general. This is why the construction of the affine group scheme \(\operatorname{Spec} A\) (for \(A\) a commutative Hopf algebra) is meaningful only when one restricts to commutative target algebras \(B\), and the resulting group is generally non-commutative as a group (but the underlying algebras \(A\) and \(B\) remain commutative).