Difficulty understanding proof of Yoneda's Lemma

Question

I've been trying to wrap my head around Yoneda's Lemma but it isn't clicking. I've been reading through this post here but I need some help.

I spent a while starring at this commutative diagram on Wikipedia:

Here is how I'm thinking about this and how it goes astray. I think the outer square is showing us the images of $A$ and $X$ under both functors, then we fix any $f : X \rightarrow Y$ and the inside square follows the behavior of $Hom(A,f)$ and $F(f)$, respectively in order to properly establish $\Phi_*$. So clearly $Hom(A,f)$ maps $id_A$ to $f$ (by definition of $Hom(A,-)$). Now we freely pick $u$ (because the theorem says each such choice leads to a unique $\Phi$). Then we know that $F(f)$ acts on $u$ by taking it to $(Ff)u$ and so we define $\Phi_X$ to take $f$ to that to get commutativity. Fine,but how does this extend to arbitrary $X,Y$ where none of them are $A$ if all we know how to do is map $id_A$? In other words, If on the topmost left corner square we replace $Hom(A,A)$ with $Hom(A,Y)$ then how do we know $\Phi_Y$ and $\Phi_X$ do the right thing, or how do we even know how they behave

Here is a diagram where I attempt to look at what happens for arbitrary $X$ and $Y$. I know I need a $g : X \rightarrow Y$ to get a map $Hom(A,g)$ which must take an $f : A \rightarrow X$ to $g \circ f \in Hom(A,Y)$. Then I know the bottom right corner is $\Phi_Y(g \circ f)$ but I don't know where $f$ goes. Taking it from the Wikipedia square (which took $f$ to $(Ff)u$ I now have something like $F[(Ff)u]$ should equal $\Phi_Y(g \circ f)$?

Answer 1

OK : point 1: it seems to me that you are confusing the map with its image: i.e., $f \ne {\rm Hom} (A, f) $; instead $f = {\rm Hom} (A, f) id$.

Point 2: you said $f$ and $f_* ={\rm Hom} (A, f)$ live in different categories - yes... a priori $f$ and $id$ are MORPHISMS in one category, and $f_*$ in another. The idea of Yoneda is to consider $f$ and $id$ as ELEMENTS of SETS, which are objects in the category (Set) where $f_*$ is a morphism: $id \in {\rm Hom} (A,A) $ and $f \in {\rm Hom} (A,B)$ so that it makes sense to write $f_*(id) =f$, i.e., $f$ is in the image of $f_*$.

Comment - as I noted in the comment sections, the above originally was meant to appear in the comment section, and not here. I am leaving it, as it allows me to correct here a mis-statement in the original, and for 'posterity.'

Note: The statement of Yoneda often comes in two 'parts' - although the proof is 'identical' in both cases. The question treats the second part.

So - for completeness - here is the first (embedding) part:

Suppose $\mathcal C$ is a category where ${\rm Hom}(A,B)$ is a set, for any two objects $A$ and $B$ - for some people (me), this is in the definition of category, but for others, such categories are called locally small.

Write ${Set}^{\mathcal C}$ to denote the category of (co-variant) functors from $\mathcal C$ to the category of sets: objects are functors, and arrows are natural transformations.

The Yoneda functor is the contravariant functor $Y\colon {\mathcal C} \rightarrow {Set}^{\mathcal C}$, which on objects carries
$$A \mapsto {\rm Hom} (A,- ),$$ and on arrows, takes $g \in {\rm Hom} (A,B )$ to

$$g^*\colon {\rm Hom} (B,- ) \rightarrow {\rm Hom} (A,- ),$$ where $g^*_C\colon {\rm Hom} (B,C ) \rightarrow {\rm Hom} (A,C )$ is the map

$$ f \mapsto f \circ g.$$ (I'm writing the subscript $C$ in $g^*_C$ to conform with the notation above.)

The 'embedding part' of the Yoneda lemma is that the (contravariant ) Yoneda functor $Y$ is faithful and full - i.e. $g\mapsto g^*$ is injective, and every arrow (natural transformation)

$$ \Phi\colon {\rm Hom} (B,- )\rightarrow {\rm Hom} (A,- )$$ is of the form $g^*$.

As I mentioned in the conversation off-line, a category theory friend of mine liked to say that the Yoneda embedding statement is a tautology, and paraphrased it by saying: "the statement of Yoneda is: an object is determined by the arrows from (or to, depending which version of Yoneda) the object. The proof of Yoneda is: in particular, the object is determined by the identity."

In any case, the proof of this version of Yoneda goes as above:

If $\Phi_B(id_B) = g \in {\rm Hom} (A,C )$, then to calculate $\Phi_C(f)$, for $f \in {\rm Hom} (B ,C )$:

Since $f =f \circ id = f_*(id_B) $,

$$ \Phi_C (f) = \Phi_C ( f_* (id_B) ) = f_*\Phi_B(id) = f_*(g) = f\circ g = g^*_C (f),$$ so $\Phi = g^*$.

For the injectivity $g^*_B(id) = g $.

Remark: consistency suggests subscripts on $f_*$ too - but I didn't in the first exchange, and it makes it harder to read anyway. Usually, one doesn't on $g^*$ either...

Answer 2

Palace Chan, not sure if you are still perplexed by Yoneda's lemma but if so I may be able to help. Looking at an early post you write: "I now have something like F[(Ff)u]should equal ΦY(g∘f)?". You appear to be treating F here like a function acting on the set F(X). The functorial notion can be confusing that way, but F does not act as a set function, taking elements of F(X) to elements of another set. It takes objects in the category A to objects in the category of sets.

I found that a good introduction to understanding Yoneda was via looking at monoids, a special type of category. Yoneda's lemma turns out to reduce to a simple algebraic fact. If you look at the action of elements of a monoid M acting on M by left multiplication, this "left action" maps the M homomorphically into the set of permutations of M. One can ask, what are the mappings of M into itself of M that commute with the left actions, ie the set of endomporhisms of M under this action? They turn out to be just the set of right actions of M. You can find the right action element of M corresponding to a homomorphism g as being just g(1). This is analogous to the Yoneda lemma's mapping of a natural mapping from Hom(A,-) to F, by finding the action of that mapping on 1A. Natural mappings are analogous to homomorphisms in the monoid case.

If interested, I can write all this out in more detail?