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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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    Since no one has linked to [this MO question](http://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical) yet...2011-09-23

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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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    Nope. Because I like $\sqcup$ and $\sqcap$ for arbitrary categories and $\oplus$ just in additive or abelian categories.2013-09-04
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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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In (modern) mathematics those morphisms which arise from the definitions or the universal properties are called the $\mathbf{canonical}$ morphisms. The functorial morphisms are called the $\mathbf{natural}$ morphisms; for instance if $\phi:F\rightarrow G$ is a natural transformation between the two functors $F,G:\mathscr{C}\rightarrow\mathscr{D}$ then for each object $C$ of $\mathscr{C}$ the morphism $\phi_{C}:F(C)\rightarrow G(C)$ is called a natural morphism. In Grothendieck's style of mathematics, all of the morphisms (which appear in practice) are either canonical or natural.