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In his book (Categories for the working mathematician) MacLane speaks (on page 45) about the category of objects (of $\textbf{Ab}$) under $\mathbb{Z}$ which is the comma category $(\mathbb{Z}\downarrow \textbf{Ab})$, and says "it is the category of abelian groups with a selected element" (in analogy with $(\star\downarrow\textbf{Set})$), but, is there not already a prefered (or selected) element in all (abelian) groups, namely the identity element?

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up vote 4 down vote accepted

Pedro, sometimes we want to pick out an element or elements as an extra structure on abelian group. For example, we may be interested in a specified $\mathbb{Z}$-basis if the abelian group has one. Comma category constructions give a way of talking about such extra structure.

Of course, as you say there is already a distinguished element given by the identity. This is reflected in the fact that there is a canonical functor

$$Ab \simeq (\{0\} \downarrow Ab) \to (\mathbb{Z} \downarrow Ab)$$

obtained by postcomposing with $\mathbb{Z} \to \{0\}$. The comma category under $\mathbb{Z}$ gives a way of picking out other elements, if we want.

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The category $(\mathbb{Z} \downarrow \mathbf{Ab})$ is by definition the category whose elements are the pairs $\langle f, G \rangle$, where $G$ is an abelian group and $f \colon \mathbb{Z} \to G$ is a group homomorphism.

Hence the selected element is $f(1) \in G$. Of course it is the identity of $G$ if and only if $f$ is the trivial homomorphism.

Ps. Saunders is the first name, the family name is Mac Lane.

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I seem to have a memory that the space in "Mac Lane" was something he introduced after a typesetter's mistake, and that his daughter has since removed the space. – Todd Trimble Jul 11 '11 at 12:43
Is it true that giving only $f(1)$ is sufficient to characterize the group homomrphism $f$? – Pedro Jul 11 '11 at 13:07
Pedro: yes, certainly. That's basically what people mean when they say that $\mathbb{Z}$ (equipped with the element $1$) is the free abelian group generated by a single element. – Todd Trimble Jul 11 '11 at 13:36
Perfect, thanks – Pedro Jul 11 '11 at 15:40

Here is an easy example which illustrates Francesco Polizzi's answer: Consider the morphisms $\mathbb{Z} \rightarrow\mathbb{Z}/12$ and $\mathbb{Z} \rightarrow\mathbb{Z}/6$, sending 1 to the class of 3. These are objects in the comma category. Now the canonical surjection $\mathbb{Z}/12 \rightarrow \mathbb{Z}/6$ induces a morphism in the comma category (it sends the class of 3 to the class of 3).

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