2
$\begingroup$

I am clumsily trying to read my way through Algebra: Chapter 0. Among the first examples of categories presented in the text is the following. Let $\mathcal{C}$ be a category and fix one of its objects $A$. Consider then Ob($\mathcal{C}_A$) to be the morphisms $f$ from any object of $\mathcal{C}$ to $A$. At this point the author suggests there is only one sensible way of defining morphisms in this category, which is: given $Z_1$, $Z_2$ in $\mathcal{C}$ and $f_1:Z_1 \to A$, $f_2:Z_2 \to A$, morphisms are the functions $g$ s.t. $f_1=g \circ f_2$ i.e. the obvious commutative diagram commutes. However, It isn't clear to me where exactly this example would fail the definition of category if we simply ignored the requirement that $g$ make the diagram commutative, i.e. if we let $g$ be any function $Z_1 \to Z_2$. I agree it wouldn't be sensible, and I'm probably missing something very easy, but still...

Thanks in advance

$\endgroup$

2 Answers 2

2
$\begingroup$

Your example would not fail the definition of category -- it would be a category. However, such category is not very interesting -- it is simply equivalent to the full subcategory of $\mathcal{C}$ with objects which have arrows to $A$. The unnaturality of this construction is obvious: every two morphisms $f_1,f_2\colon Z\to A$ from the same object are isomorphic by the identity arrow $id(Z)$ by this definition.

$\endgroup$
2
$\begingroup$

This just gives you the subcategory of $\mathcal C$ containing those objects which map to $A$-the maps themselves aren't involved, the basic principle is whenever you have structured objects, the morphisms between them should preserve that structure. I would say this should be familiar from abstract algebra, but you're reading an unusual book which does categories in general before giving you the natural examples, so you probably just have to trust that this will eventually be more intuitive.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .