What does "Arrows are more important than objects" really mean?

Question

I am a final year undergraduate student and I am trying to learn category theory. I am familiar with the basic notions. I am reading Pareigis's notes, http://www.mathematik.uni-muenchen.de/~pareigis/Vorlesungen/04SS/Cats1.pdf.

In his introduction, he states " Category theory proves that all information about a mathematical object can also be drawn from the knowledge of all structure preserving maps into this object. The knowledge of the maps is equivalent to the knowledge of the interior structure of an object. “Functions are everywhere!” "

I am not sure what he means exactly by this. I have seen before sentences like "It is the arrows that matter most". But I don't see the big picture I guess. Can someone give me some example of how the arrows, in some circumstance, give all the information about the object in a category? (It might be easier for me to understand examples coming from group theory or topology.)

Familiar with 'categories without objects'? Have a look at http://www.tac.mta.ca/tac/reprints/articles/17/tr17.pdf page 39. Note that a functor $F$ can be described as a pair $(F_o,F_a)$ where $F_o$ denotes its objectfunction and $F_a$ its arrowfunction. Essential is that $F_o$ is completely determined by $F_a$, so actually you don't need it. Also the object of a category can be identified with the arrow that serve as its identity. — drhab, Feb 01 '14 at 10:37
As a general rule (that works for me at least), when I open a book for the first time, I do not waste too much time reading the introduction. After describing the intended audience of the book, most authors summarize the main results of the text or suggest possible learning paths, but they generally write this in a somewhat technical language that readers completely ignorant of the contents of the book find difficult to understand.... — magma, Feb 01 '14 at 13:56
.... So, generally, after quickly browsing the introduction , I jump to the first chapter, which is certainly a must-read. After reading some chapters, I reread the intro, and at that point most of it makes sense. Instead of stopping at the end of page 3 in Pareigis' book, you should have proceeded up to page 14 line 7 where Pareigis explains the concept in a way that is understandable by a reader who has followed the book up to that point. — magma, Feb 01 '14 at 14:01

score 10 · Answer 1 · edited Apr 13 '17 at 12:58

10

Well, objects without morphisms are essentially boring. Think of finite-dimensional vector spaces over a field $K$. They are all essentially just $K^n$. So why study them at all? Well, the linear maps between them are interesting! One would like to classify them.

Morphisms serve as "communication" between objects. The theme "arrows are more important than objects" is best illustrated in full generality by the Yoneda Lemma. But actually there are a lot of familiar and specific examples in mathematics.

Algebra: The fundamental theorem on homomorphisms is the tool to work with quotient groups (as well as quotient rings, etc.). It is not really important that they are made up out of cosets, but rather that homomorphisms $G/N \to H$ correspond to homomorphisms $G \to H$ which "kill" $N$. Actually this is the whole idea of this construction: We want to kill elements.

Algebraic geometry: Let $f \in \mathbb{Z}[x,y]$ be a polynomial in two variables, for example $f(x,y) = x^2 - 2 y^2$. Then the solutions of $f$ in a commutative ring $R$ are exactly the homomorphisms of rings $\mathbb{Z}[x,y]/(f) \to R$.

Differential geometry: One tries to understand a manifold $M$ by means of its vector bundles, which are maps $V \to M$ (with extra structure).

Algebraic topology: One tries to understand a nice space $X$ by means of its homotopy groups, which are made up out of maps $S^n \to X$ in the homotopy category. These are groups because $S^n$ carries a natural cogroup structure.

Combinatorics: A coloring of a set $X$ with $n$ colors is just a map $X \to \{1,2,\dotsc,n\}$.

Representation theory: We try to understand a group $G$ by its $K$-linear representations, which are homomorphisms $G \to \mathrm{GL}_n(K)$.

The list is endless, therefore I will stop here. It is very easy to find examples, because almost every modern mathematical publication promotes the theme.

You might be also interested in the theme of "categorical characterizations", started by Freyd, Bergman and others (see for instance here or there or here).

edited Apr 13 '17 at 12:58

Community

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answered Feb 01 '14 at 10:50

Martin Brandenburg

181,922

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Ahh starting to make much more sense now. So, and correct me if I am wrong, the theme does not say "objects are useless" but rather, that "the arrows can be used to extract information about the object". And also, I guess it also says, "IN A CATEGORY, the arrows are more important than the objects", in that, once you have fixed a category, the arrows can be used to give information about the object; not that a group, for example, in its own existence, is not an important object. Am I correct in this? – mare_nnoem Feb 01 '14 at 11:27
I agree with you. – Martin Brandenburg Feb 01 '14 at 11:36
Thanks, very helpful. Is it true, by the way, that all of your examples are instances of the Yoneda Lemma? – mare_nnoem Feb 01 '14 at 11:40
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In some sense, yes. The Yoneda Lemma is a positve result: Every object $X$ is determined by its hom-functor $\hom(X,-)$ (or $\hom(-,X)$). Here we compare $X$ with all other objects of the category. In specific examples, e.g. mentioned above, this is not always possible. We restrict the class of "test" objects. For example homotopy groups only refer to spheres. It is a nontrivial theorem by Whitehead that a map between CW-complexes, inducing isomorphisms on homotopy groups, is already a homotopy equivalence. This can be seen as an improvement of the Yoneda Lemma for the homotopy category. – Martin Brandenburg Feb 01 '14 at 11:44
Thanks again. Please allow me to ask this, though. Excuse me if this does not make sense, I am trying to understand what is going on here. Let $\mathbf{Grp}$ be the category of groups. Then for a group $G$, the Yoneda Lemma says, if I am not mistaken, that for a functor $F:\mathbf{Grp}^{op}\to \mathbf{Set}$, there is an isomorphism $F(G)\cong \mathbf{Hom}{[\mathbf{Grp}^{op},\mathbf{Set}]}(\mathbf{Hom}{\mathbf{Grp}}(-, G),F)$. To have a result about THE group $G$, shouldn't I get an isomorphism of the form "$G\cong ...$"? Should I take $F$ to be a particular functor to see the "theme"? – mare_nnoem Feb 01 '14 at 12:06
A corollary of the Yoneda Lemma is that $\hom(-,X) \cong \hom(-,Y)$ implies $X \cong Y$ (for objects $X,Y$ of any category). – Martin Brandenburg Feb 01 '14 at 12:12
Ah, of course. Thanks again. Very helpful. – mare_nnoem Feb 01 '14 at 12:16
Excuse me for coming back to this. I am trying to put everything in my head in a correct way. The Yoneda Lemma says that every object can be described by its hom-functor, that's why "arrows matter more". So it is Yoneda that gives meaning to the theme. But...In the case of a group, we have $G\cong \mathbf{Hom}_{\mathbf{Grp}}(\mathbb{Z},G)$, so considering group homomorphisms we retrieve $G$. But this is not Yoneda, is it? It is arrows, though... – mare_nnoem Feb 01 '14 at 14:38
Also, do the examples you have given above, come for free? Following a general tool of category theory, if they are not Yoneda? I mean, it is not apparent to me, how, categorically, one can describe the group $G$ using its $K$-linear representations, for example. What is the categorical relation between the object and the morphisms $G\to \mathbf{GL}_n(K)$? And where does it come from? Does it follow from a general principle or method? – mare_nnoem Feb 01 '14 at 14:40
$G \cong \hom(\mathbb{Z},G)$ is not correct. If $U(G)$ is the underlying set of $G$, we have $U(G) \cong \hom(\mathbb{Z},G)$. Notice that $G$ is much more than $U(G)$. In order to recover $G$, we must take into acount the cogroup structure on the group $\mathbb{Z}$. Notice that $U(G) \cong \hom(\mathbb{Z},G)$ is not a corollary of Yoneda, but Yoneda tells you that $\mathbb{Z}$ is uniquely determined by this property! It is the free group on one generator. 2. This is a little bit too vague. Maybe you think over it and ask this in a new question - let us avoid chatting here.

Martin Brandenburg

Feb 01 '14 at 14:44

I agree with Martin, and would just like to add the importance of proving a particular construction is a limit or colimit by verifying the universal property, instead of checking that a particular construction works. Another virtue of the "arrows" approach is the analogies it gives between different categories: the constructions of, say, colimits may be quite different but they are still colimits. All this contributes to the unity of mathematics. – Ronnie Brown Feb 02 '14 at 10:33

score 4 · Answer 2 · answered Feb 01 '14 at 10:56

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As @Adeel mentions in the comments, learn the Yoneda Lemma, which is the actual answer to your question.

For an example from topology, consider the quotient space $X/\sim$ where $\sim$ is an equivalence relation. Do you want to think about the topology on that?

If you're like me, then no, you don't, but the good news is that there's no need to -- you can just use that a continuous map out of $X/\sim$ is the same as a map out of $X$ with $f(x) = f(y)$ whenever $x \sim y$. Yoneda's lemma is what guarantees you that you won't miss out on anything about $X/\sim$ by thinking of it this way (you still need the topological construction to know that it exists, but if you trust your textbook, you don't need to use that construction).

(You could say something similar about quotient groups -- if you like, you can think of them in terms of cosets, but you never have to do this to prove anything interesting about them.)

I've motivated the "arrows are more important than objects" viewpoint from the perspective of laziness, but I should mention two caveats:

Somehow the "right"/"most elegant" definition of, say, a quotient object is the functorial one, and the particular construction in a given category is not so important.
For more complicated objects like moduli spaces, where the construction might take a whole book, this stops being a question of laziness and starts being a question of practicality.

answered Feb 01 '14 at 10:56

hunter

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+1. It is not true that we don't need explicit constructions at all. For example, when $Y$ is a locally compact space, then the canonical bijective continuous map $(X \times Y) / (\sim \times Y) \to X / \sim \times Y$ is a homeomorphism. This fact is used quite often in topology, especially when $Y=[0,1]$. In order to prove this, we really need to work with the open subsets. This is a nontrivial statement, which cannot be derived from universal properties (we interchange certain limits with certain colimits here). – Martin Brandenburg Feb 01 '14 at 11:48
interesting example. I guess sometimes you have to look under the hood ! – hunter Feb 01 '14 at 12:15
@MartinBrandenburg what do you mean by $(\sim \times Y)$ on the lhs ? it seems a typo – magma Feb 01 '14 at 15:14
he just means to extend $\sim$ to $X \times Y$ trivially on $Y$, i.e. glue $(x_1, y_1)$ to $(x_2, y_2)$ iff $x_1 \sim x_2$ and $y_1 = y_2$. – hunter Feb 01 '14 at 15:15
That's right. Perhaps somebody knows a better notation? – Martin Brandenburg Feb 01 '14 at 15:17
@MartinBrandenburg I never encountered this notation before, that's why I asked. Is it standard? – magma Feb 01 '14 at 15:23

Giorgio Mossa · Answer 3 · 2014-02-01T12:05:08.240

There are many different answer to this question I'm gonna write some of the them.

For start we can observe that when dealing with structured set (for instance groups or topological spaces) lot's of the set theoretical structure can be recovered respectively via group homomorphisms and contiuous functions. For instance there's a one-on-one correspondence between the elements of (the underlying set of) a group $G$ and the homomorphisms of the type $\mathbb Z \to G$. Similarly there's a one-on-one correspondence between the points of a space $X$ (i.e. the elements of the underlying set) and continuous function of the form $\{*\} \to X$ (where $\{*\}$ is the pointed space with the obvious topology).

So we can indentify elements of the underlying sets of these structure with the some arrows in their respective categories, meaning that we can recover the element-structure from arrows.

Another example (which is still a topological one is the following): consider the topological space $\{0,1\}$ with the topology $\{\emptyset,\{0,1\},\{1\}\}$ and let's call this space $\Omega$. There's a one-on-one correspondence between continuous functions of type $X \to \Omega$ and open sets of $X$ (for every $f \colon X \to \Omega$ we have the open set $f^{-1}(\{1\})$ and for every $A \subseteq X$ open we have the continuous function $f_A \colon X \to \Omega$ sending every $x \in A$ in $1$ and the other points in $0$).

So again we could recover some internal structure of the topological space (the open sets) in an arrow theoretic fashion.

Finally there's the above mentioned yoneda lemma which proves that morphisms/arrows are what really matters.

In every category $\mathbf C$ for every $c \in \mathbf C$ there is a canoical functor $\mathbf C(-,c) \colon \mathbf C^\text{op} \to \mathbf {Set}$. This functor associate to every $x \in \mathbf C$ the set of morphism $\mathbf C(x,c)$. Yoneda lemma states that for every other $c' \in \mathbf C$ the two functor $\mathbf C(-,c)$ and $\mathbf C(-,c')$ are naturally isomorphic (i.e. there's a family of natural bijections $\langle \mathbf C(x,c) \to \mathbf C(x,c')\rangle_{x \in \mathbf C}$) if and only if $c$ and $c'$ are isomorphic as objects of $\mathbf C$.

This says that the structure (up to isomorphism) of the object of a category can be recovered by the families of sets $\langle \mathbf C(x,c)\rangle_{x \in \mathbf C}$, so by the arrows on that objects.

Of course there are other good example of how to recover information of an object from the arrows but I hope this ones could be sufficient to motivate the motto morphisms are what matters the most.

Thanks. Are the results in your first (second, really) paragraph, about groups and topological spaces, consequences of the Yoneda Lemma? If yes, may I ask how? ... — mare_nnoem, Feb 01 '14 at 13:57
@mare_nnoem I don't think so. Here why: yoneda lemma holds for every category while the reduction of points to special kind of arrows is something that holds for groups and topological (and other special kind of concrete categories). To be more detailed the point-arrow fact holds in those concrete categories (i.e. categories with a faithful $\mathbf{Set}$-valued functor) the underlying-set functor is representable (and that's not always the case). — Giorgio Mossa, Feb 01 '14 at 14:06

What does "Arrows are more important than objects" really mean?

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