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I mean Endofunctor maps category $A$ to category $A$ - so all of it's objects, and morphisms must remain the same. Isn't this an identity Functor?

For example, if I have a category $A$ of integers with an ordering on them:

$O = \{1,2,3,4\}$
$M(x,y) = \{(x,y) | x,y \in O, x \leq y)\}$

any Endofunctor I try to think of ends up being an identity Functor...

What I am thinking wrong here? Thanks.

EDIT: As per suggestion, let me try mapping all objects, via and endofunctor $F$ to $1$ and all morphisms to $1 \leq 1$.

$$F_{O}: O_{A} \to O_{A}, F_M: M_{A} \to M_{A}$$

then $F_{O}(1) = 1, F(2) = 1, ...$ and $F_{M}( M(1,2) ) = M(1,1) $

so now either:

  1. Nothing is changed and $A$ remains the same (endofunctor having only "mapped" things around)
  2. Or the new $A$ is now missing all objects except for $1$ and all morphisms except for $1\leq 1$.

If it's the former (1), what is the point of this functor in the first place, if it does not "do" anything...

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2  
In your example, try mapping every object to $1$ and every morphism to $1\leq 1$. –  Dejan Govc Nov 15 '12 at 17:11
    
But doesn't that change the set of "objects"? Now we only have one object! I thought category $A$ is only equal to another category $A$ if all objects and morphisms are the same? –  drozzy Nov 15 '12 at 17:38
    
I imagine you meant to write "$F_O(2)=1$" rather than "$F(2)=2$"? –  Brad Nov 15 '12 at 21:19
    
Regarding your point 2, there isn't a "new $A$". The functor $F_O$ you have defined maps $A$ to itself: for each object $a$ of $A$, $F_O(a)$ is an object of $A$, and if $f:a\to b$ is a morphism of $A$ then $F_O(f):F_O(a)\to F_O(b)$ is a morphism of $A$. It is true that the "image" of this functor is the subcategory of $A$ that has only the object $1$, but that's okay. –  Brad Nov 15 '12 at 21:22
1  
By analogy, an endofunction of a set $X$ is just a function $f:X\to X$. So, for example, the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2$ is an endofunction of $\mathbb{R}$. The image of this function doesn't contain, for example, $-1$, but it is still a function $\mathbb{R}\to\mathbb{R}$. And it is not the identity function $id(x)=x$ either. –  Brad Nov 15 '12 at 21:24

4 Answers 4

up vote 3 down vote accepted

You seem to give extra constraints to the definition of "endofunctor". In reality it is just a functor from/to the same category. It does not have to be an identity.In particular, note that an endofunctor does not have to be full or faithful or injective/surjective on objects. Consider - as a parallel - a set $A$ and a function $f: A \to A $. Is $f$ necessarily the identity function? Of course not! It could be a permutation on set A or a constant function, or...

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For the same reason, every "endomap" of sets $f: X \longrightarrow X$ would always be an identity, wouldn't it?

So, for instance, the only function $f: \mathbb{R} \longrightarrow \mathbb{R}$ would be the identity function, $f(x) = x$. And things like $f(x) = x^2, \sin x , e^x \dots$ and the like strange, inexplicable phenomena. :-)

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Take the category of "data structures", it's objects are basic types like numbers $\mathbb N$, booleans $\mathbb B$, pairs of objects $(a,b)$ and lists of objects $[a]$.

An example of a functor is the list functor: if we have a morphism like $\text{odd} : \mathbb N \longrightarrow \mathbb B$ then the list functor gives us $[\text{odd}] : [\mathbb N] \longrightarrow [\mathbb B]$ (which computes [F,T,F,T] from [0,1,2,3])

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No. The objects and morphisms don't have to remain the same. An endofunctor is just a functor of the form $A\to A$, for a category $A$.

For example, $X\mapsto \{0,1\}\times X\ $ is an endofunctor on $Set$.

In your example also any constant mapping will give an endofunctor, also ($1\mapsto 1,\ 2\mapsto 1,\ 3\mapsto 4,\ 4\mapsto 4$) works...

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I'm sure the OP would appreciate an example of a non-identity endofunctor on some category. –  Cameron Buie Nov 15 '12 at 17:10
    
But... that changes the set of objects in the category, wouldn't that change the category $A$ to something like $A'$? –  drozzy Nov 15 '12 at 17:39
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@drozzy: Why would it? The other objects are still there, even if the functor isn't mapping anything to them. –  camccann Nov 15 '12 at 18:09
    
@camccann Perhaps I don't know what $X\mapsto \{0,1\}\times X$ means... could you explain? –  drozzy Nov 15 '12 at 18:35
    
@drozzy: I was talking about the constant mapping example, but the same applies to both. Perhaps a more straightforward example would be an ordering on the natural numbers, and a functor that maps each number to its successor? –  camccann Nov 15 '12 at 18:53

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