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Buschi Sergio
Buschi Sergio
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Is very easy prove that $(Set, \otimes, 1)$$(Set, \times, 1)$ is monoidal (by elements checking). Now let $\mathcal{C}$ a category by finite product $\times$ and (then) with a final object $1$. Consider the axioms of monoidal category for $(\mathcal{C}, \times , 1)$ stated by diagrams (see for example p.462 of "Closed Categories" by Eilenberg & Kelly, LA Jolla 1967), now it remains to prove that these diagrams are commutative. COnsider a such diagram $\textbf{D}$ and a (general) object $X\in \mathcal{C}$ and the representable $(X, -): \mathcal{C}\to Set: A \mapsto (X, A)$, acting by $(X, -)$ on this diagram, we get a similar diagram in $Set$, say $X(\textbf{D})$, and $(X, -)$ preserve the product $\times$ and the final object $1$, now we just know that $(Set, \otimes, 1)$$(Set, \times, 1)$ is monoidal, then $X(\textbf{D})$ is commutative, because. Because this is true for each object $X$, by Yoneda lemma follow that $\textbf{D}$ is commutative (more easily observe that given $f, g: A \to B$ such that, if $(X, f)=(X, g): (X, A)\to (X, B) $ for each $X$ then $f=g$ (consider $X=A$ and $1_A$)).

Is very easy prove that $(Set, \otimes, 1)$ is monoidal (by elements checking). Now let $\mathcal{C}$ a category by finite product $\times$ and (then) with a final object $1$. Consider the axioms of monoidal category for $(\mathcal{C}, \times , 1)$ stated by diagrams (see for example p.462 of "Closed Categories" by Eilenberg & Kelly, LA Jolla 1967), now it remains to prove that these diagrams are commutative. COnsider a such diagram $\textbf{D}$ and a (general) object $X\in \mathcal{C}$ and the representable $(X, -): \mathcal{C}\to Set: A \mapsto (X, A)$, acting by $(X, -)$ on this diagram, we get a similar diagram in $Set$, say $X(\textbf{D})$, and $(X, -)$ preserve the product $\times$ and the final object $1$, now we just know that $(Set, \otimes, 1)$ is monoidal, then $X(\textbf{D})$ is commutative, because this is true for each object $X$ by Yoneda lemma follow that $\textbf{D}$ is commutative (more easily observe that given $f, g: A \to B$ such that $(X, f)=(X, g): (X, A)\to (X, B) $ for each $X$ then $f=g$ (consider $X=A$ and $1_A$)).

Is very easy prove that $(Set, \times, 1)$ is monoidal (by elements checking). Now let $\mathcal{C}$ a category by finite product $\times$ and (then) with a final object $1$. Consider the axioms of monoidal category for $(\mathcal{C}, \times , 1)$ stated by diagrams (see for example p.462 of "Closed Categories" by Eilenberg & Kelly, LA Jolla 1967), now it remains to prove that these diagrams are commutative. COnsider a such diagram $\textbf{D}$ and a (general) object $X\in \mathcal{C}$ and the representable $(X, -): \mathcal{C}\to Set: A \mapsto (X, A)$, acting by $(X, -)$ on this diagram, we get a similar diagram in $Set$, say $X(\textbf{D})$, and $(X, -)$ preserve the product $\times$ and the final object $1$, now we just know that $(Set, \times, 1)$ is monoidal, then $X(\textbf{D})$ is commutative. Because this is true for each object $X$, by Yoneda lemma follow that $\textbf{D}$ is commutative (more easily observe that given $f, g: A \to B$, if $(X, f)=(X, g): (X, A)\to (X, B) $ for each $X$ then $f=g$ (consider $X=A$ and $1_A$)).

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Buschi Sergio
Buschi Sergio
  • 4.2k
  • 1
  • 22
  • 26

Is very easy prove that $(Set, \otimes, 1)$ is monoidal (by elements checking). Now let $\mathcal{C}$ a category by finite product $\times$ and (then) with a final object $1$. Consider the axioms of monoidal category for $(\mathcal{C}, \times , 1)$ stated by diagrams (see for example p.462 of "Closed Categories" by Eilenberg & Kelly, LA Jolla 1967), now it remains to prove that these diagrams are commutative. COnsider a such diagram $\textbf{D}$ and a (general) object $X\in \mathcal{C}$ and the representable $(X, -): \mathcal{C}\to Set: A \mapsto (X, A)$, acting by $(X, -)$ on this diagram, we get a similar diagram in $Set$, say $X(\textbf{D})$, and $(X, -)$ preserve the product $\times$ and the final object $1$, now we just know that $(Set, \otimes, 1)$ is monoidal, then $X(\textbf{D})$ is commutative, because this is true for each object $X$ by Yoneda lemma follow that $\textbf{D}$ is commutative (more easily observe that given $f, g: A \to B$ such that $(X, f)=(X, g): (X, A)\to (X, B) $ for each $X$ then $f=g$ (consider $X=A$ and $1_A$)).