# Is every functor monoidal between monoidal categories where monoidal product is interpreted as sum?

Namely, for every functor $F$, is $(F, (A_0, A_1)\mapsto[F(\iota_0(A_0, A_1)), F(\iota_1(A_0, A_1))])$ monoidal? (For reference, $\iota_i(A_0, A_1):A_i\to A_0+A_1$ is an injection of the categorical sum $A_0+A_1$.)

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If by "categorical sum" you mean "coproduct" and by "monoidal" you mean "lax monoidal", the answer is "yes".

Let's denote

$$\iota_A : A \longrightarrow A \sqcup B \longleftarrow B : \iota_B$$

the natural arrows. Applying any functor $F$ to the diagram above, we get

$$F\iota_A : FA \longrightarrow F(A \sqcup B) \longleftarrow FB : F\iota_B \ .$$

But we also have the same diagram for the coproduct of $FA$ and $FB$:

$$\iota_{FA} : FA \longrightarrow FA \sqcup FB \longleftarrow FB : \iota_{FB}$$

Hence, by the universal property of the coproduct, morphisms $F\iota_A$ and $F\iota_B$ induce a unique arrow

$$FA \sqcup FB \longrightarrow F(A\sqcup B ) \ .$$

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Sorry, Agustí, you just repeated my construction. You did not prove axioms of monoidal functor. –  beroal Sep 21 '11 at 9:35
All the axioms should follow from the uniqueness of the arrow. –  a.r. Sep 21 '11 at 12:13
This is your proof? Ha-ha-ha. Низачот. –  beroal Sep 30 '11 at 8:07
Sorry you didn't like it. Sure enough you'll be able to do better. –  a.r. Sep 30 '11 at 10:00
I have to correct my former statement: "All the axioms should follow from the uniqueness of the arrow" [$FA\sqcup FB \rightarrow F(A\sqcup B)$]. I'm afraid it was ambiguous. I should have said: "All the axioms follow from the uniqueness of the arrow." Hope it is more clear now. –  a.r. Sep 30 '11 at 18:23