# In the category of Abelian Groups, what precisely is the canonical morphism? What about for the category of sets?

I would like a relatively concrete answer. Thanks!

Sorry, I need to be more precise. I am trying to show that the category of Sets is distributive. So, given any sets $X, Y,$ and $Z$, I want to know what the canonical morphism is from $(X\times Y)\oplus(X\times Z)\to X\times(Y\oplus Z)$.

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What's the "canonical morphism" ? Morphisms are between 2 objects in a category, and usually there are many morphisms among the objects in a category (certainly for Abelian Groups and Sets). – Ted Sep 22 '11 at 6:15
I have tried to add $\LaTeX$ code. I am not 100% about $\oplus$, the OP means co-product, which Wikipedia denotes as $\oplus$. Please confirm/correct this use! – Asaf Karagila Sep 22 '11 at 6:25
For sets we could certainly use $\coprod$. – Dylan Moreland Sep 22 '11 at 6:30
– Dylan Moreland Sep 22 '11 at 6:33
By functoriality of $\times$ and the property of $\coprod$, you have maps $\text{id}_X\times i_Y\colon X\times Y\to X\times(Y\coprod Z)$ and $\text{id}_X\times i_Z\colon X\times Z\to X\times(Y\coprod Z)$. These factor through $(X\times Y)\coprod(X\times Z)$. – Yuri Sulyma Sep 22 '11 at 6:39

Yuri Delanghe's comment is correct, of course, but I am not sure it satisfies your desire for concreteness. In the case of sets, $(X\times Y)\oplus(X\times Z)$ is the disjoint union of $X\times Y$ and $X\times Z$. A standard way of making two sets disjoint is multiplying the first set with $\{0\}$ and the second with $\{1\}$. We use the same technique for $Y\oplus Z$. Now each element of $(X\times Y)\oplus(X\times Z)$ is of the form $((x,a),i)$ where $a\in Y\cup Z$ and $i\in\{0,1\}$. The canonical morphism maps this to $(x,(a,i))$, which is in $X\times(Y\oplus Z)$.
In the case of abelian groups, $X\times(Y\oplus Z)$ and $(X\times Y)\oplus(X\times Z)$ are not isomorphic, in general. There is a nice morphism from $X\times(Y\oplus Z)$ to $(X\times Y)\oplus(X\times Z)$, though. Namely, since for finitely many factors product and coproduct coinside with the direct sum of abelian groups, the elements of $X\times(Y\oplus Z)$ are of the form $(x,(y,z))$ with $x\in X$, $y\in Y$, and $z\in Z$. Such an element gets mapped to $((x,y),(x,z))$ in $(X\times Y)\oplus(X\times Z)$.