Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
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
1  
@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 3

up vote 2 down vote accepted

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 $1_A$)).

share|improve this answer
    
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.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.