My question is about the notion of exponential functors as they are frequently defined in the literature on (strict) polynomial functors, e.g., the paper "General Linear and Functor Cohomology" by Franjou, Friedlander, Scorichenko, and Suslin, or the paper "Bar complexes and extensions of classical exponential functors" by Touze. The essentials of the definition are as follows:

Let $k$ be a field, and let $A$ be a functor from the category of finite-dimensional (f.d.) k-vector spaces to the category of k-vector spaces. The tensor product of two such functors is defined in the obvious way.

Given such a functor $A$, suppose there exists a natural transformations $k \rightarrow A$ (viewing $k$ here as a constant functor) and $m: A \otimes A \rightarrow A$ such that for each f.d. vector space $V$, $A(V)$ is made into an associative $k$-algebra. Now say that $A$ is an *exponential functor* if for each pair of f.d. vector spaces $V$ and $W$, the composite map

$A(V) \otimes A(W) \stackrel{A(\iota_V) \otimes A(\iota_W)}{\rightarrow} A(V \oplus W) \otimes A(V \oplus W) \rightarrow A(V \oplus W)$,

induced by the inclusions $\iota_V: V \rightarrow V \oplus W$ and $\iota_W: W \rightarrow V \oplus W$ and the multiplication in $A(V \oplus W)$, is an isomorphism of $k$-vector spaces. Examples of exponential functors include the functor $S$ that takes a vector space $V$ to the symmetric algebra $S(V)$, the functor $\Lambda$ that takes $V$ to the exterior algebra $\Lambda(V)$, or the functor $\Gamma$ that takes $V$ to the divided power algebra $\Gamma(V)$.

Now, it seems to be well-known in the area that if $A$ is an exponential functor in the above sense, then each $A(V)$ is automatically not just an algebra, but a $k$-bialgebra, with the coproduct defined as the composite map

$\Delta_A(V): A(V) \rightarrow A(V \oplus V) \rightarrow A(V) \otimes A(V)$,

where the first arrow is the map induced by the diagonal map $V \rightarrow V \oplus V$, and the second arrow is the exponential isomorphism $A(V \oplus V) \cong A(V) \otimes A(V)$. Details are provided in the literature for why $\Delta_A(V)$ is always coassociative, but never (that I have found) for why $\Delta_A(V)$ satisfies the remaining condition for $A(V)$ to be a bialgebra, i.e., $\Delta_A(V)$ should be an algebra homomorphism.

Why is $\Delta_A(V)$, as defined above, always an algebra homomorphism? Or, is it always an algebra homomorphism? What conditions on $A$ suffice to make $\Delta_A(V)$ always be an algebra homomorphism?

My question is a bit vague, because I have not told you how to view the tensor product $A(V) \otimes A(V)$ as an algebra; for the examples $A = S$ and $A = \Gamma$, it is the ordinary tensor product of algebras, while for $A = \Lambda$ it is the graded tensor product of algebras (the grading coming from the natural grading on $\Lambda$). So in general, the answer should probably depend on some kind of grading inherent to $A$.

One easy case where I can see that $\Delta_A(V)$ should be an algebra homomorphism is when each $A(V)$ is a (graded) commutative algebra, and $A(V) \otimes A(V)$ is viewed as the orindary (graded) tensor product of algebras, for then $\Delta_A(V)$ can be seen to be the composite of two algebra homomorphisms.