Monoidal structure on a category with products and with terminal object

Let $K$ be a category with products $(X,Y)\mapsto X\sqcap Y$ and with a terminal object $T$. It seems obvious to me that $\sqcap$ and $T$ define a structure of a monoidal category on $K$, but I can't find a reference. When I try to prove this myself I come to amazingly bulky constructions. Is there a text where this is accurately proved, or at least formulated?

-
I don't know a reference, but I do know this has a name: ncatlab.org/nlab/show/cartesian+monoidal+category . – Eric Peterson Nov 1 '12 at 19:41
Eric, thank you, that's interesting. So this means that the accurate proof exists... It would be nice to look at it... – Sergei Akbarov Nov 1 '12 at 20:00
Is easy (elements check) that $Set$ is cartesian (i.e. for finite products) monoidal, then for a general cartesian category you apply the (general) representable $(X, -)$ to the axioms diagrams (and use the result in $Set$), then the commutativity of each diagrams follow from Yoneda Lemma (need only the faithful part). – Buschi Sergio Nov 1 '12 at 20:03
Sergio, I don't understand this trick. Is it possible, for example, to prove the diagram of associativity (the pentagon) in this way? – Sergei Akbarov Nov 1 '12 at 20:20
@Sergei, using the Yoneda lemma it is, because that tells you the functor $X\mapsto Hom(-,X)$ is fully faithful. – David Roberts Nov 1 '12 at 23:15

3 Answers

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 $1_A$)).

-
Interesting... OK, I need a time to verify the details. – Sergei Akbarov Nov 2 '12 at 7:54
It would amount to the same thing if we thought about this in terms of generalized elements. – Spice the Bird Nov 2 '12 at 17:03
Generalized elements? What's this? Anyway, I accept Sergio's answer. – Sergei Akbarov Nov 2 '12 at 17:36
Let $\mathcal{C}$ be a category and $X$ be an object. Then a generalized element of $X$ of shape $Y$ is a morphism, $f:Y\rightarrow X$. We may also write this as $f\in_{Y} X$, for $f$ is a $Y$ shaped element of $X$. See ncatlab.org/nlab/show/generalized+element and also see the book at the website patryshev.com/books/Sets%20for%20Mathematics.pdf. – Spice the Bird Nov 3 '12 at 2:30

You can find it as an example of a monoidal category in Tom Leinster's "Higher Operads, Higher Categories", which contains loads of coherence proofs for higher categories.

-
I didn't understand, is this proved there, or just formulated? – Sergei Akbarov Nov 1 '12 at 20:06
Just formulated, but the book gives a lot of information that will allow you to prove it yourself. – Wouter Stekelenburg Nov 1 '12 at 23:17
Thank you, I'll try to find this book. – Sergei Akbarov Nov 2 '12 at 7:55

You need to have chosen products for every pair of objects using the axiom of choice, otherwise you don't get a product functor, just a product anafunctor and I recommend you don't try to use those just yet. Then you can prove - and this is the key step - that any two bracketings of an iterated product are isomorphic in a unique way when you demand the isomorphism respects the all the projections. The unique such isomorphism for a triple product is then the associator, and the uniqueness of the isomorphism for the 4-fold product means that the pentagon commutes. Ditto with the other coherence conditions.

Note that you can choose the product of any object $X$ with the terminal object to be $X$. This makes the unit conditions automatic.

-
David, yes, I agree that the operation $(X,Y)\mapsto X\sqcap Y$ must be a mapping. Suppose this is so, then you say that the Yoneda lemma allows to simplify the proof? Can you recommend a text where this trick is used (not necessarily for proving what I am asking about, but just for something...), I would like to look how this trick works. – Sergei Akbarov Nov 2 '12 at 6:25