What does Yoneda’s Lemma mean?
As I promised, today I’ll try to explain what Yoneda’s Lemma really does for us.
First off, let’s think about functions of two variables. To be very explicit, think of the function 
defined by 
. We can turn this into a function of one variable that returns another function. Specifically, if we write 
for the set of functions from the natural numbers to themselves, we can write 
, where =xy](https://s-ssl.wordpress.com/latex.php?latex=%5Cleft%5B%5Coverline%7Bf%7D%28x%29%5Cright%5D%28y%29%3Dxy&bg=e6e6e6&fg=333333&s=0 "\left[\overline{f}(x)\right](y)=xy")
. That is, 
is the function that sends 
to 
. This method of “Currying” (named for Haskell Curry) a function should be familiar if you’ve worked with functional programming languages like LISP, ML, or Haskell. There’s really a lot to be said about it in general, and I’ll get back to exploring it in more depth another time.
Anyhow, a similar statement goes for functors of more than one variable. Since such functors are “functorial in each variable”, we can stick an object into all but one slot and be left with a functor of the last slot. The thing is, this assignment is itself functorial.
Let’s focus on a functor 
. Then for any object 
we have a functor 
. Now when we change the object 
by applying a morphism 
we get a natural transformation 
. It should have components 
. The clear choice is 
, and indeed this is natural since the morphisms from 
commute with those from 
.
Now, strictly speaking this 
may not be a functor since we only know that 
is a category when at least one of and 
is small, but the general idea should be clear.
Now to the situation at hand: we know that any category comes equipped with a functor 
. We apply this Currying operation on each slot to get two “functors” 
and 
. Of course, the targets are only sure to be categories when (and thus 
as well) is small, but we’ll gloss over that point. What the Yoneda Lemma says is that these functors are fully faithful!
To see this, notice that 
sends an object 
to the covariant functor 
, while 
sends an object 
to the contravariant functor 
. Since the objects of and are the same, I hope that using for both won’t be too confusing.
Now let’s take the statement of the (covariant) Yoneda Lemma and apply it to the functor 
for some object 
. It says that 
. But notice that 
is 
— the set of morphisms from 
to in the functor category 
. Now since is a functor from to , it acts on the hom sets, sending 
to 
. And the Yoneda Lemma tells us that this is a bijection — the functor is fully faithful!
And the same argument goes through for . Now we apply the contravariant Yoneda Lemma to see that 
. The functor induces a bijection on each hom set, so it’s fully faithful too!
Now remember what the point of fully faithful functors is: they’re all about full subcategories. This means that any (small) category can be seen as a full subcategory of 
. And full subcategories are nice because they just restrict the objects we’re interested in without throwing out any morphisms between those objects. So, the objects of any category can be seen as a special kind of contravariant 
-valued functor on that category.
And now: which contravariant -valued functors on a category are so special that they constitute objects of under this embedding? Well, as usual in category theory we’re interested in things up to isomorphism, so we should ask, “What contravariant set-valued functors are (naturally) isomorphic to one in the image of ?”
These are just the representable functors.
The representable, contravariant, -valued functors on a category constitute a full subcategory of all contravariant -valued functors on , which is equivalent to the category itself.
And that’s why Yoneda’s Lemma is so amazing.
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June 7, 2007 -
Posted by John Armstrong |
Category theory
Shouldn’t the requirement in par 4 by that $\mathcal C$ and therefore $\mathcal C^{\bf op}$ be locally small? (c.f. the ‘size worries’ paragraph Tom Leinster’s The Yoneda Lemma: What’s it All About.
How much trouble do reasonably adequate math grad students have with this thing? I’ve read it described as ‘the hardest trivial thing in mathematics’, which makes me feel better about having so much trouble with it. This and its predecessor post are however a very helpful presentation, perhaps due to having a ‘clean plot line’ without a lot of extraneous stuff
The lack of a preview or edit comment feature on wordpress is really getting to me, here’s an attempt to fix the link:
The Yoneda Lemma: What’s it All About. Well might as well try to get the latex right too, for once:
what does Sub-categories mean!!!!!!
☺
michelle: I mentioned that in my post on full and faithful functors.
MathOutsider: The average graduate student in algebra, algebraic topology, or algebraic geometry can understand the content of the Yoneda Lemma. To them, it is a generalization of Cayley’s theory for groups and the fact that a left R-module M is isomorphic to Hom(R, M) (viewed as an R-module).
On the other hand, unless the study is able to take a course that specifically covers category theory, they are unlikely to be taught the Yoneda Lemma in any particular course.
Marceau, I’m not so sure that’s the case. Maybe I would have said that back when I was in graduate school, but as I’ve seen more and more graduate students I really have to revise that “average” statement. Particularly the applied mathematics students who don’t really have as thorough a background in algebra would have a huge problem with understanding Yoneda without an explicit explanation.
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\mathrm{Nat}(h_A,h_B)$ – I guess this is a latex error…
No, I just left off the “$latex” opening. good catch.
“we only know that \mathcal{C}^\mathcal{B} is a functor”
Shouldn’t this be “is a category”?
I’ve probably missed something obvious, but I can’t see why we have that y is fully faithfull follows from the fact that there _exists_ a natural isomorphism between Hom(A, B) and Hom(y A, y B), why does this say anything about how y acts on Hom(A, B)?
The point is that
‘s action on the hom spaces actually is the natural isomorphism. It’s not just that the two are naturally isomorphic, but that the isomorphism is given in this particular way.
So the fact doesn’t follow from the statement of the Yoneda lemma so much as the specific isomorphism in this particular proof?
If this is the case I find the presentation given here, and in most other places, somewhat strange, since they seem to imply that the fact follows from the statement of the Yoneda lemma (which I’ve never seen stating isomorphism explicitly).
The introductory chapter of “Introduction to higher order categorical logic” is the only place I’ve seen this made explicit.
Yes, the traditional statement just asserts existence, but the proof is very explicit. You have a point that some people may be confused (evidently you have been) but that’s part of why I go into the meaning and highlight the upshot of the proof.
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