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](http://arxiv.org/abs/math/9909194)" by Franjou, Friedlander, Scorichenko, and Suslin, or the paper "[Bar complexes and extensions of classical exponential functors](http://arxiv.org/abs/1012.2724)" 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?
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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.



When is an exponential functor a bialgebra?
In this situation one can use the language of monoidal categories. The terminology which I use below is standard in category theory.


Tensor porduct of functors between categories $\mathcal{V} \rightarrow \mathcal{W}$ can be defined as soon as $\mathcal{W}$ is a monoidal category. The functor category $[\mathcal{V}, \mathcal{W}]$ becomes a monoidal category under this tensor. A monoid $A$ in this monoidal category is the same as a functor $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$, where $\mathrm{Mon}(\mathcal{W})$ is the category of monoids in $\mathcal{W}$. 


If $\mathcal{W}$ is symmetric (or just braided) monoidal category then $\mathrm{Mon}(\mathcal{W})$ becomes a monoidal category itself. In this case a bimonoid in $\mathcal{W}$ is defined to be a comonoid object in $\mathrm{Mon}(\mathcal{W})$. If $\mathcal{V}$ itself is monoidal and $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ is an opmonoidal functor (this means that there are certain structural maps $A(X\otimes'Y) \rightarrow A(X)\otimes A(Y)$), then $A$ will take comonoids in $\mathcal{V}$ to comonoids in $\mathrm{Mon}(\mathcal{W})$, i.e. to bimonoids. 


Suppose that $\mathcal{V}$ is monoidal under the direct sum. Then the functor $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ becomes monoidal (a notion dual to opmonoidal) with the structural maps as defined in the question. If the monoidality structure morphisms are invertible, then $A$ also is opmonoidal with the inverse maps as the structural morphisms. 


Every object of $\mathcal{V}$ has a canonical comonoid structure via the diagonal. Under the conditions explained $A : \mathcal{V} \rightarrow \mathrm{Mon}(\mathcal{W})$ will take these canonical comonoids to bimonoids in $\mathcal{W}$.


**Edit:** This to work, the monoidal structure maps (the exponential maps) have to be actually monoid morphisms. This holds when $A(X) \rightarrow A(X\oplus Y)$ and $A(Y) \rightarrow A(X\oplus Y)$ commute with each other in the monoid structure of $A(X\oplus Y)$ in the appropriate sense. That is what is pointed out in the comments to OP.


