Lax monoidal functor Let me denote $Cat$ the category of small categories. It is a symmetric monoidal category with respect to the cartesian product. Let $F: (Cat, \times)\rightarrow (Set,\times)$ a symmetric monoidal functor functor.
Suppose that $g: C\rightarrow D$ a monoidal lax monoidal functor between small monoidal categories.
I was wondering if
$$ F(g): F(C)\rightarrow F(D)$$ is a multiplicative morphism between monoids?
 A: Not necessarily; however, if $F$ is a $2$-functor (so it sends natural transformations to identities), then the answer is yes.
If you weaken it to $F$ sending naturally isomorphic functors to equal arrows, then the answer is no in general, but yes if $g$ is strong monoidal.
If you don't have any assumption of that form, the answer is just no, in fact $F$ might not even send $C$ (and $D$) to a monoid ! (unless you require strict monoidality)
Examples :

*

*For the last one, consider the functor $F = Ob$ sending a category to its set of objects. Then $F(C) = Ob(C)$ isn't even a monoid if $C$ isn't strict monoidal.


*For the second one, consider $ F = \pi_0$ sending a category to its set of isomorphism classes. Note that if $g_0,g_1 : C\to D$ are naturally isomorphic functors, then $\pi_0(g_0) = \pi_0(g_1)$. It follows that if $C$ is monoidal, then $\pi_0(C)$ is a monoid. However, if $g: C\to D$ is only lax monoidal, e.g. $1_D\to g(1_C)$ is not an isomorphism, then one can show that $g(1_C)$ is not isomorphic to $1_D$ (even abstractly) and so $\pi_0(g)$ doesn't send the unit to the unit. Take for example $C=$ a skeleton of the category of finite dimensional $\mathbb F_p$-vector spaces with the tensor product and $D=$ a skeleton of the category of finite sets with the cartesian product; and $g : C\to D$ the forgetful functor with its usual lax monoidal structure. In this example, neither the unit nor the multiplication are respected by $\pi_0(g)$.
If $F$ sends any natural transformation to identities, then clearly $F(C),F(D)$ are monoids and furthermore the lax monoidal structure on $g$ amounts to a square which commutes up to a possibly noninvertible natural transformation :
$$\require{AMScd}\begin{CD}C\times C @>>> D\times D \\
@VVV @VVV \\
C @>>> D\end{CD}$$
as well as a triangle which again commutes up to a possibly noninvertible transformation, given by the units. If $g$ is strong monoidal, then these diagrams commute up to an invertible transformation, and so the weaker hypothesis is sufficient. As explained above, this is optimal.
