Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,
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?

share|improve this question
add comment

3 Answers 3

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.