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Let $A,B,C$ be objects of a category of modules over a ring. It is not hard to see that the Yoneda embedding "reflects exactness" (as Weibel puts it, on p. 28), i.e. if

$\hom(X,A)\stackrel{f_*}{\to}\hom(X,B) \stackrel{g_*}{\to} \hom(X,C)$ is an exact sequence of abelian groups for every module $X$, then $A\stackrel{f}{\to}B\stackrel{g}{\to}C$ is an exact sequence of modules.

This is done in Weibel, or in this article of Paul Garrett.

There, he states that this is a special case of the Yoneda lemma.

But I'm struggling to see this is the case. Here is my (naive) attempt:

$\ker(g_*)=\ker(\hom(A,g))\simeq \hom(A,\ker(g))$ because the covariant Hom functor is preserves kernels (because it preserves limits, because it has a left adjoint)

I would very much like to have $\operatorname{im}(f_*)\simeq \hom(A,\operatorname{im}(f))\ \ \ \ (*) \ \ $ in the same way; if I had that, the given exactness would get me $\hom(A,\operatorname{im}(f))\simeq \hom(A,\ker(g))$. And then I would be happy, for Yoneda lemma would instantly get me $\operatorname{im}(f)\simeq \ker(g)$.

But I'm not happy because I can't see why $(*)$ holds. It's false that Hom preserves images.

However, I'm baffled because if this is supposed to be an application of Yoneda lemma, then surely it must be because $\hom(A,\operatorname{im}(f))\simeq \hom(A,\ker(g))$... or perhaps not?

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Ah, I'm sure this is an easily answered question. I feel a bit dumb posting it. However, my extreme lack of proficiency with Yoneda and his gadgets suggests me I might not come up with a clean answer myself. – Bruno Stonek Jan 13 '12 at 22:19
(There is an \hom and a \ker, giving $\hom$ and a $\ker$, respectively :) ) – Mariano Suárez-Alvarez Jan 13 '12 at 22:36
@Mariano: Ah, indeed, I didn't know that. Pity there's not an \im! – Bruno Stonek Jan 13 '12 at 23:13
up vote 11 down vote accepted

(Thanks for the ping, @Pierre-Yves Gaillard...)

As in Pierre-Yves Gaillard's answer, I think that the usual formalization of "Yoneda's Lemma" does not literally yield the exactness assertion quoted. However, if one tolerates a certain amount of figurative reference, it's not completely ridiculous to refer to using $X\rightarrow Hom(X,-)$, and inferences about the original category from the image, as Yoneda-ish. After all, isn't that why we care about the Yoneda maps?

As Pierre-Yves G notes, and as in @Agusti Roig's elaboration of the key point of the argument, there are further ingredients in play, such as additivity or abelian-ness, which were built into the example scenario my note treated, so I did not have to name them. Thus, the pure Yoneda Lemma is not directly addressing all the issues. However, the idea to "consider" $X\rightarrow Hom(X,-)$ is the watershed, I think. If we take the simplest illustrative situation, abelian groups, so that ("by luck") the Hom(,)'s are in the same category, etc., one can immediately proceed to "do the thing" with no formalism whatsoever. One of the nice didactic points is that we really, really do need the "naturality" to know that the squares commute, to reach the conclusion. This is the simplest illustration I know (apart from the usual algebraic topology examples) of the on-the-street genuine content of "naturality", as opposed to the all-too-common cocktail-party misunderstanding of it as "not making arbitrary choices", etc.

But, yes, then the narrative was not connected to any formalism, either. The features of $X\rightarrow Hom(X,-)$ that made things work were not abstracted or formalized.

I'd plead guilty to "formal incorrectness", but with extenuating circumstances, as follows. Namely, the point of the linked-to note was to illustrate the immediate utility of $X\rightarrow Hom(X,-)$ to very tangible issues, such as showing right-exactness of $-\otimes X$, by seeing that it is a left adjoint, and showing left adjoints (with additivity...) are right exact, for very general reasons. The intended audience had/has little prior acquaintance with category theory or homological algebra, and that note was intended as an advertisement or promotion of such ideas, without formalizing those ideas enough to name them precisely. If I recall correctly, in particular, someone had been wrangling with a direct argument for the right exactness of tensor-product, and had gotten bogged-down in details that shouldn't matter, and I was trying to make the case to them that the causality in the situation could (without paying too heavy a price) be better understood as being about adjoint functors, etc. To make that case, there was absolutely no room for any set-up. At the same time, the name-dropping seemed appropriate on philosophical grounds.

I am in no sense "a category theorist" or "homological algebraist", but do have some appreciation for the widespread usefulness of the ideas, "even" without formalism or abstraction of categorical notions. Indeed, I would claim that the ideas "are there", whether one recognizes/formalizes them or not, and that recognizing them is immediately useful in the most pragmatic sense. Here I am playing against a common belief that one "chooses", or not, to "do" category theory or homological algebra, and that ignoring those ideas is a viable, completely sensible option.

In that context, perhaps over-selling "Yoneda" by throwing in a few further ingredients is an excess that can be forgiven? (Not quite as bad as a "stone soup" scenario.)

I do also admit to an ever-waning interest in formalization, especially definitions. Pithy examples seem to me much better, even without a pre-existing name for the phenomenon illustrated. One "problem" with this is that naturally-occurring illustrative examples often are not "pure", in the sense that more than the single phenomenon is involved. E.g., if I say I'm illustrating Yoneda by the exactness argument at hand, the illustration is not "pure", and there is potential confusion, indeed. On the other hand, if I'm trying to convince someone of the virtues of an idea (the $X\rightarrow Hom(X,-)$), it would be unwise to deliberately ignore the "extra" features they actually cared about.

In summary: indeed, Yoneda does not literally entail that exactness conclusion, I think. Nor need it be invoked in the simple case at hand. Rather some aspects of a special case of Yoneda are being proved directly, along with some additional bits to address the problem at hand.

Edit: in response to @Bruno Stonek's follow-up question, the "naturality" is the obvious, essentially trivial assertion that exactness of the top row entails exactness of the bottom row in the following: $$ \matrix{ 0 & \rightarrow & Hom(LX,A) & \rightarrow & Hom(LX,B) & \rightarrow & Hom(LX,C) \cr & & \downarrow & & \downarrow & & \downarrow \cr 0 & \rightarrow & Hom(X,RA) & \rightarrow & Hom(X,RB) & \rightarrow & Hom(X,RC) } $$ where the vertical arrows are isomorphisms, and $L,R$ are left and right adjoint to each other. That is, without naturality (or whatever name one likes), the vertical isomorphisms would not necessarily assure that the squares commute.

Another Edit: in response to Bruno Stonek's further "tangential" question... The "additivity" or some similar qualifier is necessary for "exactness" to make sense at all. In familiar concrete categories where we'd have any impulse to mention it, such as categories of modules over some ring, it's just structure we're used-to. Not every category behaves this way, of course. Sufficient notions to talk about kernels and cokernels can obviously be axiomatized...

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It's perfectly clear now, thank you very much for the detailed answer. I understand your reasons to call it a special case of Yoneda lemma, in that it is an illustration of Yoneda reasoning, and it certainly has the same flavor. – Bruno Stonek Jan 14 '12 at 21:39
I've reread your answer and I have a question: where exactly is the illustration of "naturality" you speak of in the third paragraph? – Bruno Stonek Jan 14 '12 at 21:59
Ah, I thought you meant that, but as it wasn't clear to me from your formulation that you were addressing the proof that adjoint additive functors are side exact,I wanted to make sure, in case you meant another illustration of naturality :) By the way, just so you know: if you ping someone in an answer, it doesn't alert the person. At least, this didn't ping me... I saw the edit by chance. – Bruno Stonek Jan 14 '12 at 22:39
@Bruno Stonek, ... thanks for the point about "pinging"... – paul garrett Jan 14 '12 at 22:41
Sorry to bother you again, but there's something (tangential to the question) that's been bothering me. You say "left adjoints (with additivity...) are right exact". But I fail to see where additivity enters the proof! Would you clarify this for me? Thank you very much – Bruno Stonek Jan 20 '12 at 20:18

I don't know if you can use more explicitly Yoneda to prove this, but if in

$$ \hom(X,A) \stackrel{f_*}{\longrightarrow} \hom(X,B) \stackrel{g_*}{\longrightarrow} \hom(X,C) $$

you take

  • $X = A$, you'll get that $0=g_*f_*(\mathrm{id}_A) = gf$. Hence, $\mathrm{im}\ f \subset \ker g$.
  • $X = \ker g$, and consider the inclusion $\iota \in \hom (\ker g, B) $, you'll have that $g_*(\iota) = g\iota = 0$. Hence, there is some $\kappa \in \hom (\ker g, A)$ such that $\iota = f_*(\kappa) = f\kappa$. And so, $\ker g = \mathrm{im}\ \iota = \mathrm{im}\ (f\kappa) \subset \mathrm{im}\ f$.
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Yes, I understand this, it's also the proof given both by Weibel and Garrett. However, Garrett in the article linked in the OP says that this result is a special case of Yoneda lemma; but I don't see how this is so, or how the proof uses it. – Bruno Stonek Jan 14 '12 at 3:22

Here is how I understand Paul Garrett's comment in the text linked in Bruno's question.

For any category $C$, let $C'$ be the opposite category. Let $A$ be an abelian category and $G$ the category of abelian groups. "Functor" shall mean "additive functor".

The following statements are straightforward.

$\bullet$ The category of functors from $A$ to $G$ is abelian in a natural way.

$\bullet$ The natural functor from $A$ to $\hom(A',G)$ is full, faithful, and left exact.

$\bullet$ Let $F:A\to B$ be a full and faithful left exact functor between abelian categories, let $$a_1\to a_2\to a_3$$ be arrows in $A$, and assume that $$Fa_1\to Fa_2\to Fa_3$$ is exact. Then $$a_1\to a_2\to a_3$$ is exact.

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Thank you for your answer. But it's not clear to me how this involves the Yoneda lemma... – Bruno Stonek Jan 14 '12 at 15:01
Dear @Bruno: I pinged Paul Garrett here, and I noticed that he was l last seen 13 hours ago. I hope he'll answer your question. - My understanding is that the statement doesn't follow immediately from YL, in the technical sense of the term. But it follows from it in a kind of philosophical way, because you think of an object as "being" the functor it represents. Also the statements in my answer are in some way an "abelian category version of YL". - I'll try to think more about it... Thanks for your nice question! (+1) – Pierre-Yves Gaillard Jan 14 '12 at 15:58
Thank you very much for your help! – Bruno Stonek Jan 14 '12 at 16:46

Yoneda Lemma, In my opinion, is to demonstrate Yoneda embedding functor (which is fully-faithful).

You can prove without any difficulty to show that if an embedding functor is exact at the same time, then "it reflect the exactness" (By homology calculation).

So statement and proof you menstioned in Weibel's textbook just means that exactness of this embedding is not necessary to "reflect the exactness".

Here is the statment:

Assume two abelian categories $\mathscr{A}$, and $\mathscr{B}$, and an addictive fuctor $E$ between them. In additioin, $E$ is fully-faithful, then we have:

Consider $a \xrightarrow{f} b\xrightarrow{g} c$ in category $\mathscr{A}$, if $E(a) \xrightarrow{E(f)} E(b)\xrightarrow{E(g)} E(c)$ is exact in $\mathscr{B}$, then $a \xrightarrow{f} b\xrightarrow{g} c$ if exact in $\mathscr{A}$. you take

  • $E(g)E(f)=E(gf)=0\Rightarrow gf=0$ implies $\mathrm{im}(f)\subseteq \mathrm{ker}(g)$.

  • consider the inclusion $\iota \in \hom (\ker g, b)\cong\hom(E(\ker g),E(b)) $, by following exact sequence: $$ 0\rightarrow\hom(E(\mathrm{ker}g),E(a)) \rightarrow \hom(E(\mathrm{ker}g),E(b))\rightarrow \hom(E(\mathrm{ker}g),E(c)) . $$ we can find $\kappa \in \hom (\ker g, a)$ such that $\iota = f\kappa$, And so, $\ker g = \mathrm{im}\ \iota = \mathrm{im}\ (f\kappa) \subset \mathrm{im}\ f$.

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