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I went to a lecture on category theory, and I am trying to understand something: I want to know what it means for a category to be distributive. We were given a homework problem, but I don't even understand what it means: Let $C$ be a category such that for all $X,Y \in C$ there exists a product $X \times Y$ and a coproduct $X \oplus Y$ in $C$. Show that there is a canonical morphism

$\psi : (X \times Y) \oplus (X \times Z) \to X \times (Y \oplus Z)$. When $\psi$ is an isomorphism, $C$ is called a {distributive category}.

What does "Show there is a canonical morphism $\psi : (X \times Y) \oplus (X \times Z) \to X \times (Y \oplus Z)$" mean? What does "canonical morphism" mean? By the way, functors have not yet been introduced in the course. So far we have only learned the definition of category, product, and coproduct.

Please don't use the word "universal property" in your answer. I don't understand that yet.

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See math.stackexchange.com/questions/65170/… Are distributive categories the new black? –  Mariano Suárez-Alvarez Sep 23 '11 at 2:08
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"Canonical" does not mean anything, really, apart from "interesting to us". –  Mariano Suárez-Alvarez Sep 23 '11 at 2:09
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Please do not ask the same question here and on MO at the same time (mathoverflow.net/questions/76176/…) –  Mariano Suárez-Alvarez Sep 23 '11 at 2:10
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@Alison: Dear Alison, I know that you asked people not to use the expression "universal property", but ... : If you have seen products and coproducts, then you have seen some universal properties, and you have to use them if you are going to investigate properties of these constructions. (In an arbitrary, unspecified, category, there is no way to say anything about a product or a coproduct than via the universal properties that they satisfy.) In this context, "canonical" means a morphism that you can construct using just the universal properties of the objects involved. (In a typical ... –  Matt E Sep 23 '11 at 2:40
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Canonical is not a precise mathematical term, but rather has to do with human psychology: a morphism is canonical if any two sensible people who come up with such a morphism are extremely likely to arrive at the same one. –  Omar Antolín-Camarena Sep 23 '11 at 3:00
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2 Answers

up vote 3 down vote accepted

I've always thought that "canonical" means "unique..." having a certain (universal, sorry) property.

So, for instance, the universal property (which you can translate as "the definition") of the coproduct $A\sqcup B$ (sorry, I don't like the notation $\oplus$ for the coproduct) of two objects $A$ and $B$ is the following: by definition it's an object $A\sqcup B$ together with two morphisms

$$ A \stackrel{\iota_A}{\longrightarrow} A\sqcup B \stackrel{\iota_B}{\longleftarrow} B $$

satisfying the following property: whenever you have two morphisms

$$ f: A \longrightarrow C \qquad \text{and} \qquad g: B \longrightarrow C \ , $$

there is a canonical (unique) morphism

$$ h : A \sqcup B \longrightarrow C \qquad \text{such that} \qquad h\iota_A = f \quad \text{and} \quad h\iota_B = g \ . $$

For instance, in the category of sets, the coproduct is the disjoint union, $\iota_A$ and $\iota_B$ are the inclusions and this morphism $h$ is

$$ h(x) = \begin{cases} f(a) & \text{if}\ x=a\in A \\ g(b) & \text{if}\ x=b\in B \ . \end{cases} $$

As for the product, its universal property (definition) is: the product of two objects $A$ and $B$ is an object $A \times B$ together with two morphisms

$$ A \stackrel{\pi_A}{\longleftarrow} A\times B \stackrel{\pi_B}{\longrightarrow} B $$

satisfying the following property: whenever you have two morphisms

$$ f: C \longrightarrow A \qquad \text{and} \qquad g: C \longrightarrow B \ , $$

there is a canonical (unique) morphism

$$ h : C \longrightarrow A \times B \qquad \text{such that} \qquad \pi_A h = f \quad \text{and} \quad \pi_B h = g \ . $$

For instance, in the category of sets, the product is the usual Cartesian product, $\pi_A$ and $\pi_B$ are the projections and this morphism $h$ is

$$ h(c) = (f(c), g(c)) \ . $$

From these universal (sorry) properties follows, for instance, that whenever you have a morphism

$$ f: A \longrightarrow B $$

you can "multiply" it by an object $X$, obtaining a morphism

$$ \mathrm{id}_X \times f : X \times A \longrightarrow X \times B \ . $$

$\mathrm{id}_X \times f$ is the morphism induced by

$$ \mathrm{id}_X\pi_X : X\times A \longrightarrow X \qquad \text{and}\qquad f\pi_A : X\times A \longrightarrow B \ . $$

and the universal (sorry) property of the product. For instance, in the category of sets

$$ (\mathrm{id}_X \times f) (x,a) = (x, f(a)) \ . $$

So, your canonical morphism

$$ (X\times Y) \sqcup (X\times Z) \longrightarrow X \times (Y\sqcup Z) $$

is deduced from the universal (sorry) properties defining product and coproduct and this last construction (also deduced from the aforementioned universal properties). Namely, to begin with, you have your morphisms that come together with the coproduct

$$ Y \stackrel{\iota_Y}{\longrightarrow} Y\sqcup Z \stackrel{\iota_Z}{\longleftarrow} Z \ . $$

Next, you "multiply" them by $X$:

$$ X\times Y \stackrel{\mathrm{id}_X \times \iota_Y}{\longrightarrow}X\times ( Y\sqcup Z ) \stackrel{\mathrm{id}_X\times \iota_Z}{\longleftarrow} X\times Z \ . $$

And now, the universal property of the coproduct gives you a canonical (unique) morphism

$$ \psi : (X\times Y) \sqcup (X\times Z) \longrightarrow X \times (Y\sqcup Z) $$

such that

$$ \psi \iota_{X\times Y} = \mathrm{id}_X \times \iota_Y \qquad \text{and}\qquad \psi\iota_{X\times Z} = \mathrm{id}_X \times \iota_Z \ . $$

In the category of sets, this $\psi$ is

$$ \psi (x, a ) = \begin{cases} (x, y) & \text{if}\ (x,a)=(x,y) \in X\times Y \\ (x, z) & \text{if}\ (x,a)=(x,z) \in X\times Z \ . \end{cases} $$

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In the eight latex blob from the bottom, should it be f\pi_A instead of f\pi_B? –  Raeder Sep 23 '11 at 9:09
    
Yes. You're right. Thank you. –  a.r. Sep 23 '11 at 10:04
    
nice explanation of products & coproducts. Funny that you said sorry explicitly each time you said universal.. +1, lol –  Patrick Da Silva Dec 4 '11 at 7:20
    
Well, apparently the OP didn't like the word. :-) –  a.r. Dec 6 '11 at 4:19
    
Hey why don't you like the notation $\oplus$ for the coproduct? I'm genuinely curious. Is it because in the opposite category, it may be more reminiscent of a product? –  user18921 Sep 4 '13 at 15:11
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For a category theorist, 'canonical' does not just mean 'of interest', it has more the feeling of 'there by virtue of the general definitions giving the structures being studied'not of any particular manifestation of that context. Here in your question there is always a morphism going where you put it for the simple reason that the definitions of product and coproduct in all generality force that existence. It is 'natural' without having to make additional assumptions, or choices.The detailed argument,putting flesh on these bare bones is given by Agusti in an answer (above at present!) but the detail does mention universal properties!

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