# Category with zero

From "An Introduction to Ring Theory", Paul Cohn:

"Let $\mathcal{A}$ be any category and define $\mathcal{A}''$ as the category obtained from $A$ by adjoining one object $Z$with a single morphism $Z\rightarrow A$ and a single morphism $A\rightarrow Z$ for each object $A$ of $\mathcal{A}$, as well as a single morphism $Z\rightarrow Z$, and with the natural compositions. Verify that $\mathcal{A}''$ is a category with zero".

First of all, what's a category with zero? Does it mean a category that includes identity morphisms?

Second, what does it mean to 'adjoin morphisms to an object'? Does that mean that in $\mathcal{A}''$, the each object is a triple (an object of $\mathcal{A}$ and the two morphisms)?

• A zero object is both terminal and initial. For instance, the category of groups. – egreg Jul 3 '15 at 16:59
• Ah, so they mean it's a category with a zero object? Well, in that case isn't it pretty obvious that it's $Z$, as the two morphisms to and from it are uniquely defined? – man_in_green_shirt Jul 3 '15 at 17:02
• @man_in_green_shirt, honestly I'm not sure what there is to verify here... – goblin Jul 3 '15 at 17:06
• I don't know, looks pretty obvious now that I've understood the question – man_in_green_shirt Jul 3 '15 at 17:08

## 1 Answer

A zero object is one that is both initial and terminal, that is, admits a unique morphism to or from any other object. An example is the category of groups, where the trivial group $\{1\}$ is the zero object. The category of sets has no zero object, because the initial object is not isomorphic to any terminal object.

If $\operatorname{Obj}\mathcal{A}$ denotes the class of objects in the category $\mathcal{A}$, then $Z$ should be something such that $Z\notin\operatorname{Obj}\mathcal{A}$ and $$\operatorname{Obj}\mathcal{A}''=\operatorname{Obj}\mathcal{A}\cup\{Z\}$$ For $A,B\in\operatorname{Obj}\mathcal{A}$, we define $$\mathcal{A}''(A,B)=\mathcal{A}(A,B)$$ while $$\mathcal{A}''(A,Z)=\{0_{A,Z}\}, \qquad \mathcal{A}''(Z,B)=\{0_{Z,B}\}, \qquad \mathcal{A}''(Z,Z)=\{0_{Z,Z}\},$$ where the elements used above are pairwise distinct and don't belong to the class of morphisms in $\mathcal{A}$. Composition in $\mathcal{A}''$ is defined in an obvious way: it is the same as in $\mathcal{A}$ for morphisms already in $\mathcal{A}$, it is the unique morphism if $Z$ is involved.

Now verify that $Z$ is a terminal and initial object in $\mathcal{A}''$.