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When is it important to distinguish between an object in a category and that object's identity morphism?

I am wondering if the only reason that we consider objects at all is to avoid infinitely recursive definitions.

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up vote 5 down vote accepted

Formally, there is not. But in most "concrete" categories at hand, the objects and morphisms are quite different conceptually. This makes it natural to distinguish between them in most cases.

Furthermore, in e.g. slice categories, the identity morphism of different objects may be the same if considered as an object on its own. This requires distinguishing them in one way or another. One of the most natural ways to do so is by means of objects.

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Can you elaborate a little bit more on how identity morphisms of different objects in a slice category are equivalent? – smartcaveman Apr 25 '13 at 17:55
See this definition. The identity morphism is the same for all morphisms $f: X\to C$ with domain $X$. – Lord_Farin Apr 25 '13 at 17:57
Why is this true about slice categories? – smartcaveman Aug 9 '13 at 3:58

[I misread the question as asking why is it important to distinguish, rather than when is it important to distinguish. Even though I answered the wrong question, I liked my answer enough that I didn't want to delete it. Also, it does seem to address the "infinite regress" part of the original question. So I have marked it CW.]

Some definitions of a category don't distinguish; they drop the objects entirely.

For example, the definition of Freyd and Scedrov's Categories, Allegories (page 3) omits the objects, identifying each object with its identity morphism. If $x$ is a morphism, then $\square x$ denotes the "source" of $x$. Conceptually this is the codomain of $x$, except that the codomain object has been eliminated in favor of its identity morphism. So one has for example the axiom that $(\square x)\cdot x = x$, and the theorem that $\square(\square x) = \square x$.

Similarly $x\square$ is the "target" of $x$, the identity morphism of the codomain object of $x$, so the rule that $fg$ is defined only when the domain of $g$ and the codomain of $f$ are equal becomes "$xy$ is defined iff $x\square = \square y$".

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There are two different group structures on the two-element set $\{a,b\}$, but the identity morphism of both is simply $a\mapsto a, b\mapsto b$. Unless you add some "tagging" to your morphisms, you seem to loose information ...

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by tagging, Do you mean labels? – smartcaveman Apr 25 '13 at 17:57
I don't understand the point of this answer, and in how far it answers the question. – Martin Brandenburg Apr 25 '13 at 21:33
@MartinBrandenburg I think the point is that the category of groups contains two different objects whose identity morphisms are the same function. So if one doesn't want to distinguish objects from their identity morphisms, then one must build additional information into the morphisms. (One could, of course, get the objects to be not only different but non-isomorphic, by using a larger example.) – Andreas Blass May 2 '13 at 2:28
This is not true. The sources and targets of morphisms are not sets, not even in a concrete category: they are objects of that category. Two identity morphisms are identical iff their sources and targets are identical as objects of the ambient category, not as sets, not even in a concrete category. – Qiaochu Yuan Aug 10 '13 at 17:32

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