# What's the difference between an endofunctor and a morphism?

The question is in the title, really.

I understand the answer in a general sense: Morphisms map objects, and functors map both objects and morphisms. So an endofunctor would map morphisms as well as objects. But that's not an intuitively satisfying answer. It seems like "part" of the endofunctor is just a morphism under a different name. I'm trying to understand this at a deeper level, and the texts I have don't really cover this question, per se.

Examples might be helpful, or more clarification on the ways they are different beyond the simple version I give above. I'm not sure how to clarify what I'm after yet, but feel free to try and pin me down to specifics in comments if it helps...

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I don't really understand the question. Specializing to the case of a category with one object and all morphisms invertible (a group in disguise) we get the question "what's the difference between an endomorphism of a group and an element of a group?" and I don't know what to say to this except "one of these things is not like the other..." – Qiaochu Yuan Jul 7 '11 at 16:24
@Qiaochu, I suspect the misunderstanding is mine, rather than yours, but I'm trying to pin it down to specifics. A morphism maps an object in a category to another object in the category, right? A functor maps both objects and morphisms in one category to objects and morphisms in another category, right? An endofunctor is a functor which maps from one category back onto itself, right? So one of the things an endofunctor does is to map objects from a category onto other objects in the category, which is what a morphism does. It feels like there is overlap, but as you say, they're different. – Craig Stuntz Jul 7 '11 at 16:31
A morphism is an arrow from some specific object to some specific other object. An endofunctor contains the data of a function from the collection of objects in some category to itself. I think you should work through some examples more closely to see what the difference between these two notions is; they're on two different categorical levels. One way to say this is that a functor is a morphism between two categories in the "category of categories." – Qiaochu Yuan Jul 7 '11 at 16:35
@Qiaochu, yes, as I said, an example or two is really what I'm looking for. Your comment above is probably half of the answer, a good reference to an example or two is probably the other half. – Craig Stuntz Jul 7 '11 at 16:39

I think your confusion is, first of all, about language -so, being not a native English speaker, maybe I'm not the best qualified to help you- and different levels of abstraction that you're mixing. Secondly, maybe you lack the knowledge of examples of categories where morphisms don't map anything at all.

To begin with, maybe I could say that the meaning of your "map" in "Morphisms map objects" is not the same as in "functors map both objects and morphisms". Or shouldn't be. Or you should deffinitively avoid it in the first sentence.

In the second sentence, you're using it correctly in the categorical parlance (categorical level of abstraction), meaning "send". But this same use in the first one is not correct, because, in the categorical parlance (categorical level of abstraction) morphisms don't "map" (send) anything to anything (not necessarily, or not at all).

Outside category theory, we say (correctly) "$f(x,y) = x$ maps $\mathbb{R}^2$ onto $\mathbb{R}$", when we are doing Analysis, for instance. This is a perfectly legitimate use of the verb "to map".

But, thinking of the same $f: \mathbb{R}^2 \longrightarrow \mathbb{R}$ as a morphism -for instance, in the category of topological spaces and continuous maps- we look at it simply as an "arrow" between the two "objects" $\mathbb{R}^2$ and $\mathbb{R}$ -and you forget that it "sends" vectors $(x,y)$ to its first coordinate $x$. More correctly, we have defined what the set of continuous morphisms $\mathbf{Top} (\mathbb{R}^2, \mathbb{R})$ is and we say that $f$ is a member of this set. Full stop: we don't need anything else from the categorical point of view (categorical level of abstraction).

So, generally, you should avoid thinking that morphisms in a category map anything, meaning that they "send" something onto / into something else. You should avoid this use / meaning of the verb "to map" speaking about morphisms in a category because it's not true, correct, in general.

Instead, in the categorical parlance (categorical level of abstraction) functors do really map (send) objects to objects and maps to maps. Indeed, by definition, a functor is composed by two "functions": one that assigns objects to objects, and one that assigns maps to maps.

One example where the two uses of "map" coexist. The fundamental group functor $\pi_1$, maps (sends) topological spaces to groups and continuous maps to group homomorphisms:

$$\pi_1 : \mathbf{Top} \longrightarrow \mathbf{Groups}$$

For instance, $\pi_1$ "sends" the unit circumference $S^1$ to the group of integer numbers $\pi_1 (S^1) = \mathbb{Z}$ and the continuous map that goes round the circumference twice $\gamma : S^1 \longrightarrow S^1$, to the group homomorphisms $\gamma_ *= \pi_1 (\gamma ) : \mathbb{Z} \longrightarrow \mathbb{Z}$ which is multiplication by $2$: $\gamma_* (m) = 2m$.

But we also say, and it is true, of course, that $\gamma$ maps $S^1$ onto $S^1$. And it's also true that $\pi_1$ maps $S^1$ to $\mathbb{Z}$ (and $\gamma$ to $\gamma_*$). But they are both ($\gamma$ and $\pi_1$) different kinds of "maps", corresponding to different levels of mathematical abstraction.

Moreover, in this example with $\gamma$ and $\pi_1$, you could insist: "But, anyway, both are maps, arent't they? Both map, don't they?".

One example where they don't. Yes, but they are plenty of examples, talking about categories, where morphisms of a particular category don't map / send anything at all.

For instance, you can think of $\mathbb{Z}$ as a category, with objects the integer numbers and morphisms defined this way: we say that the set of morphisms between two (different) integer numbers $m$ and $n$ is empty if and only if $m > n$; otherwise, that is if $m < n$, we say that the set of morphisms from $m$ to $n$ has exactly one morphism $m \longrightarrow n$. (And when $m=n$, we have also just one morphism, the identity of $m$).

"Oh, but who is this morphism, when $m<n$?, which is its formula?, what does it map?", you could ask. Well, I'm sorry, but, from a categorical point of view, I don't need to tell you: it doesn't matter. My "category" $\mathbb{Z}$ is perfectly defined as it is. (You could try to be picky at this point and ask me for how does composition of morphisms work in this $\mathbb{Z}$ category, but, I don't need to say anything more because there is just one morphism for each pair $m<n$, so composition works in the only possible way.)

And a last word: you should not think that this "phenomenon" (morphisms don't map / send ) is unusual in the realm of categories

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@Craig Stuntz, @Agustí Roig: I can second that. The word “map” has different meanings in 0) “morphisms map objects” and 1) “functors map both objects and morphisms”. Categories are defined axiomatically, and their morphisms are not functions in general. Morphisms may be relations, pairs of functions. Or something cloudy, e.g. in preorder category. “Map” is used metaphorically in 0, literally in 1. P.S. I am not a native English speaker. – beroal Jul 8 '11 at 15:48
Great answer. But that describes the difference between morphisms and functors. Do you mind helping me distinguish between functors and endofunctors? – drozzy Nov 15 '12 at 19:38

An endofunctor maps objects and arrows within a category to other objects and arrows, in a functorial way. But it is not true in general that there is any morphism at all between an object and its image under the endofunctor: for example, consider the following trivial endofunctor $F : \textbf{Set} \to \textbf{Set}$, defined on objects $X$ by $FX = \emptyset$ and arrows $f$ by $F(f) = \text{id}_{\emptyset}$. It is straightforward to verify that this is indeed a functor, but there is no arrow $X \to FX$ unless $X$ is itself empty.

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