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Is very easy prove that $(Set, \otimes, times, 1)$ 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 $(Set, \otimes, times, 1)$ is monoidal, then $X(\textbf{D})$ is commutative, because . Because this is true for each object $X$ 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$1_A$)). 1 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\$)).