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 $A_0+A_1$.)
Is every functor monoidal between monoidal categories where monoidal product is interpreted as sum?
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category-theory
monoidal-categories
1 Answers
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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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0Sorry, Agustí, you just repeated my construction. You did not prove axioms of monoidal functor. – 2011-09-21
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0All the axioms should follow from the uniqueness of the arrow. – 2011-09-21
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0This is your proof? Ha-ha-ha. Низачот. – 2011-09-30
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0Sorry you didn't like it. Sure enough you'll be able to do better. – 2011-09-30
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0I 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. – 2011-09-30