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I have a very common situation, for which I need both: (1) notation; and, if available, (2) a general relative term. Let's say that:

  • there is a functor between categories, $f:C_1\to C_2$,
  • $c_1$ is a particular object of $C_1$
  • $c_2$ is a particular object of $C_2$, such that in mapping $C_1$ to $C_2$, $f$ maps $c_1$ to $c_2$

What is the name given to a morphism that maps $c_1$ to $c_2$ in the same way that $f$ does, independently of the existence of $C_1$ or $C_2$? What concise notation can I use to refer to such a morphism?

I am sure that there is a clear answer for this, but so that the point of my confusion is more clear, these are my intuitions about the issue:

  • $f$ itself can't be the answer, because it is a specialization of the morphism that I am referring to, because it communicates a lot more information than just $c_1\to c_2$. For example, it is possible that another functor could map two completely different categories and still meet the criteria of mapping $c_1$ to $c_2$. (e.g. if $g: C_3 \to C_4$ could map $c_1$ to $c_2$ despite being distinct from $C_1$ and $C_2$)

  • $f(c_1)$ seems closer to what I am looking for, but I think that $f(c_1)$ should actually refer to the resultant value, or $c_2$ itself. I am interested in the morphism between $c_1$ and $c_2$, rather than only $c_2$.

Update: If my question seems nonsensical for reasons raised by both @Jim and @AlexKruckman, let's just consider $c_1$ and $c_2$ to be categories themselves and the morphism I am asking about to also be a functor itself.

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When you say morphism mapping $c_1$ to $c_2$ is that synonymous with functor that sends $c_1$ to $c_2$? Because usually morphisms are things that are inside the category. The objects $c_1$ and $c_2$ are not in the same category. They need not be the same "type" of object and there might not be any sensible notion of what a map between them is. –  Jim Apr 25 '13 at 16:33
    
@Jim, Please see my comment under Alex Kruckman's post. –  smartcaveman Apr 25 '13 at 16:53
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3 Answers

$f$ does not provide a morphism from $c_1$ to $c_2$. In the general situation you describe, it doesn't make sense to talk about a morphism from $c_1$ to $c_2$, since they live in different categories.

You also write "a morphism that maps $c_1$ to $c_2$ in the same way that $f$ does." But $f$ just associates the object $c_1$ of $C_1$ to the object $c_2$ of $C_2$. There's no other data there - except what $f$ does to morphisms, but this depends on the categories $C_1$ and $C_2$.

If you just want to talk about the association from $c_1$ to $c_2$, independently of $C_1$ and $C_2$, you could talk about the map of sets $\{c_1\} \rightarrow \{c_2\}$, or the ordered pair $(c_1,c_2)$.

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Let's say that $c_1$ and $c_2$ are both categories themselves, thereby entailing that $c_1\to c_2$ is itself a functor, because even though $C_1\neq C_2$, $c_1$ and $c_2$ are both in ${\bf Cat}$. How does that change the answer? –  smartcaveman Apr 25 '13 at 16:45
    
@smartcaveman: So in that case, $C_1$ and $C_2$ are distinct categories whose objects are also categories, right? –  camccann Apr 25 '13 at 16:51
    
That would be correct. –  smartcaveman Apr 25 '13 at 16:52
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@smartcaveman: So in general, you're talking about relating a functor between two categories with morphisms in a third category that contains all the objects of the other two? –  camccann Apr 25 '13 at 16:55
    
@camccann, sure, except that in addition to containing all the objects of the other two categories, the third category contains the other two categories themselves (though I'm not sure this is relevant to my question) –  smartcaveman Apr 25 '13 at 16:58
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Let $\mathcal C$, $\mathcal D$ be categories.

A functor $F : \mathcal C \to \mathcal D$ is a pair of mappings:

A function $F_1 : \text{ob}(\mathcal C) \to \text{ob}(\mathcal D)$ mapping the objects of $\mathcal C$ to the objects of $\mathcal D$.

And another function $F_2$ such that given object $A,B \in \text{ob}(\mathcal C)$ and a morphism $f : A \to B$ in $\mathcal C$ we have the morphism $F_2(f) : F_1(A) \to F_1(B)$ in $\mathcal D$.

Furthermore to be a functor it must satisfy laws regarding composition and identities.

We will usually just write $F$ for either $F_1$ or $F_2$, it being decided which was referred to from context.

Given a functor $F$ you may write $F : \text{ob}(\mathcal C) \to \text{ob}(\mathcal D)$ to explicitly denote the mapping between objects.

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Is $F_1$ a function? I thought that for a morphism to be a function, its domain and codomain had to be sets, and $ob(C)$ is a class. Am I mistaken? –  smartcaveman Apr 25 '13 at 16:47
    
@smartcaveman, it is a function, ob(C) denotes the "set" of objects of the category C. –  shobon Apr 25 '13 at 16:48
    
So, can I get what I want then by using $F_2(f:c_1\to c_2)$ to get the identity morphism, and then use either the source or target operation to get the object? –  smartcaveman Apr 25 '13 at 16:52
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1. The issue of whether $F_1$ is a function depends on your choice of foundation for category theory. If ob(C) must be a set, then $F_1$ is a plain old function. If you allow ob(C) to be a proper class, then $F_1$ is a "class function", i.e. a proper class all of whose elements are ordered pairs, associating each element of ob(C) to an element of ob(D). –  Alex Kruckman Apr 25 '13 at 16:54
    
2. You write $f: c_1 \rightarrow c_2$, but it doesn't make sense to talk about an arrow from $c_1$ to $c_2$ because they live in different categories. What's true is that $F_2(id_{c_1}) = id_{c_2}$, but I think this is irrelevant to your question. –  Alex Kruckman Apr 25 '13 at 16:56
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Maybe an example would help clear this up. Do you know what the fundamental group of a topological space is?

There is the category of pointed topological spaces $\mathtt{top}_\ast$ whose objects are pairs $(X, x)$ with $X$ a topological space and $x \in X$ a point. The morphisms $(X, x) \to (Y, y)$ are continuous maps $f\colon X \to Y$ such that $f(x) = y$.

There is the category of groups $\mathtt{grp}$ whose objects are groups and whose morphisms are group homomorphisms.

The fundamental group $\pi_1$ can be thought of as a functor $\pi_1\colon\mathtt{top}_\ast \to \mathtt{grp}$ which sends a pair $(X, x)$ to the set of loops in $X$ that start/end at $x$ (actually it's equivalence classes of loops up to homotopy, this forms a group that you can read about in the above link).

Now, for example, if $S^1$ is the circle and $x \in S^1$ then $\pi_1(S^1, x) = \mathbb Z$ (as an additive group). What's important here is that this is not a map from the circle to $\mathbb Z$. Given a point in the circle there's no way to specify what loop that point should go to, there's no map!

It can be confusing because up to this point you've always studied maps between things, but remember a functor that sends $c_1$ to $c_2$ is not a map $c_1 \to c_2$!

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Thanks. I'm going to read up/think on this and get back later, because right now "a functor that is not a map" is not processing –  smartcaveman Apr 25 '13 at 17:38
    
@smartcaveman, a functor is not a map, it's a pair of mappings [that satisfies some relations] as I said in my answer –  shobon Apr 25 '13 at 18:03
    
@shobon, How can we generalize this to a functor between n-categories? Would it just be a set of $n+1$ mappings (that satisfy some relations)? –  smartcaveman Apr 29 '13 at 15:54
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