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    • CommentRowNumber1.
    • CommentAuthorjoe.hannon
    • CommentTimeJun 1st 2013
    • (edited Jun 3rd 2013)

    In reduced homology#relation to relative homology, at the bottom is a computation where one step is that kerH 0(ε)cokerH 0(x).\ker H_0(\epsilon) \cong \operatorname{coker}{H_0(x)}. Can you explain that step to me? If you think of kernels and cokernels as morphisms, then there’s no way such an isomorphism can hold, since kerH 0(ε)\ker H_0(\epsilon) is a morphism into H 0(X)H_0(X), while cokerH 0(x)\operatorname{coker}{H_0(x)} is a morphism out of H 0(X).H_0(X). But if you think of kernels and cokernels as objects, then I guess it’s possible for them to be isomorphic. My guess is it must follow from the fact that H 0(ε)H 0(x) H_0(\epsilon)\circ H_0(x) is an iso, so the statement is something like “the kernel of a split epimorphism is the cokernel of its right-inverse,” but I can’t figure it out.

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 3rd 2013
    • (edited Jun 3rd 2013)

    Your guesses are correct (of course –as you surmised – it is a very standard abuse to refer to kernels, cokernels by their representing objects, omitting the arrows that witness their universality). But here are some details (probably overkill, but I just want to make the general situation clear).

    The general situation is where we have a commutative triangle

    A f B g C\array{ A & \stackrel{f}{\to} & B \\ & \mathllap{\simeq} \searrow & \downarrow \mathrlap{g} \\ & & C }

    where we want to show coker(f)ker(g)coker(f) \cong ker(g). It is not hard to convince oneself that we may assume without loss of generality that the isomorphism is an identity, where we have a retraction pair

    A i B 1 A r A\array{ A & \stackrel{i}{\to} & B \\ & \mathllap{1_A} \searrow & \downarrow \mathrlap{r} \\ & & A }

    Let p=ir:BBp = i \circ r: B \to B; we have p 2=pp^2 = p since ri=1 Ar \circ i = 1_A. As subobjects, we can identify i:ABi: A \to B with the inclusion im(p)Bim(p) \hookrightarrow B; similarly, as quotient objects, we can identify r:BAr: B \to A with the projection Bim(p)B \to im(p). Now put q=1 Bpq = 1_B - p. Then we have pq=qp=0p \circ q = q \circ p = 0, and p+q=1 Bp + q = 1_B, and also q 2=qq^2 = q. Also note that if q(b)=0q(b) = 0, then b=p(b)b = p(b), so bim(p)b \in im(p).

    Let jj denote the subobject inclusion im(q)Bim(q) \hookrightarrow B and let ss denote the quotient object projection Bim(q)B \to im(q) taking bb to q(b)q(b). Then (j,s)(j, s) is a retraction pair (sjs \circ j is the identity on im(q)im(q)). Putting all this together, we have an exact sequence

    0im(p)iBsim(q)00 \to im(p) \stackrel{i}{\to} B \stackrel{s}{\to} im(q) \to 0

    which, by the symmetric relationship between pp and qq, is split by the exact sequence

    0im(p)rBjim(q)00 \leftarrow im(p) \stackrel{r}{\leftarrow} B \stackrel{j}{\leftarrow} im(q) \leftarrow 0

    so that im(q)im(q) is canonically identified with coker(i)coker(i) from the first exact sequence, and with ker(r)ker(r) from the second exact sequence.

    As you can see, this is an exercise in the yoga of idempotents in AbAb-enriched categories, as discussed in articles scattered around the nLab (idempotent completion, Cauchy completion, etc.).

    • CommentRowNumber3.
    • CommentAuthorjoe.hannon
    • CommentTimeJun 5th 2013
    • (edited Jun 5th 2013)

    Thank you, Todd. So the basic idea is that when the idempotent splits, we can reduce to the case of a direct sum of two objects, and isomorphism of the cokernel of inclusion of one summand with the kernel of projection onto that summand holds.

    I got a little nervous when you started working with im(p)\text{im}(p). I know three definitions of image of a map qq, kernel of cokernel of qq, smallest subobject of codomain through which qq factors, and the set-theoretic definition. I think I know they’re all equivalent in an abelian category. Do we have to assume that here?

    I also followed the links you gave and read this lovely sentence at split idempotent: if an idempotent splits, its retract is both the equalizer and coequalizer of the parallel pair of the idempotent and the identity. That fact was fun and easy to check, and seems to be related to the result I want. Cause then eq(p,1)=ker(1p)=kerq\text{eq}(p,1)=\ker (1-p)=\ker q and so kerq=cokerq.\ker q = \operatorname{coker}q. And if q=jsq=js with sj=1sj=1, then ss is epi and jj is mono, so kerq=kers\ker q=\ker s and cokerq=cokerj.\operatorname{coker}q=\operatorname{coker}j. Well we wanted to show that kerr=cokeri,\ker r=\operatorname{coker}i, not kers=cokerj\ker s=\operatorname{coker}j, but just turn the crank again with those arrows. But I’m still unsure about what category-theoretic properties the objects im(p)\operatorname{im}(p) and im(q)\operatorname{im}(q) are assumed to have, and what axioms we need to get them.

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 5th 2013

    Didn’t mean to cause nervousness! I didn’t know what generality you wanted to work in, but since the original question was about abelian groups, I used “image” in the humble sense that is familiar from one’s first experiences with homomorphisms of abelian groups, not concerning myself very much with what generality my argument would work in. If pressed on that, I would have been content to quote the abelian category axioms, whereby the image is (by definition) the cokernel of the kernel, and the canonical map from the image to the coimage is an isomorphism. Abelian categories being a good environment in which to do homological algebra.

    But seeing that you picked up the scent (and seeing also that I said at the end this was really about idempotents in AbAb-enriched categories), it is right and good to aim for that level of generality for this particular proposition, which you now have done. Taking it from the top one more time: form the idempotent p=irp = i r and its complementary idempotent q=1pq = 1 - p. Split the idempotent qq. The splitting q=jsq = j s can be realized either by taking jeq(q,1)=ker(p)=ker(r)j \coloneqq eq(q, 1) = ker(p) = ker(r), or scoeq(q,1)=coker(p)=coker(i)s \coloneqq coeq(q, 1) = coker(p) = coker(i). (One should stop for a moment and check the last equalities in those two equation-strings, but the first follows from the observation that pf=0p f = 0 iff irf=0i r f = 0 iff rf=0r f = 0 since ii is monic, and similarly the second follows from the fact that rr is epic.) Thus we have that the domain of ker(r)ker(r) is identified with the codomain of coker(i)coker(i).

    • CommentRowNumber5.
    • CommentAuthorTobyBartels
    • CommentTimeJun 5th 2013

    the image is (by definition) the cokernel of the kernel

    Techinicality: it's the kernel of the cokernel.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 5th 2013

    kernel of the cokernel

    Irk. I’ll have to remember to keep that straight.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeJun 6th 2013

    kernel of the cokernel

    I’ve never understood why. In what examples do the kernel of the cokernel and the cokernel of the kernel not agree and it makes sense to call the one the image and the other the coimage (as opposed to vice versa)? In an abelian category, they’re all the same. In the category of (nonabelian) groups, the cokernel of the kernel seems like the most natural definition of “image”, since it is also the set-theoretic image — the kernel of the cokernel is (I guess) instead its normal closure. And in the very nonabelian case of a topos or a regular category, again the correct notion of “image” is the coequalizer of the kernel pair, not the equalizer of the cokernel pair.

    • CommentRowNumber8.
    • CommentAuthorZhen Lin
    • CommentTimeJun 6th 2013

    In the case of topological abelian groups, the kernel of the cokernel and the cokernel of the kernel agree as ordinary abelian groups but may have different topologies.

    • CommentRowNumber9.
    • CommentAuthorTobyBartels
    • CommentTimeJun 6th 2013

    In what examples do the kernel of the cokernel and the cokernel of the kernel not agree

    Even in a discrete abelian setting, they pretty much never agree; they are morphisms, not objects. Specifically, the map XfYX \overset{f}\to Y factors as

    XcoimfIimfY. X \overset{coim f}\to I \overset{im f}\to Y .

    People will sometimes replace II here with an isomorphism, and that is an unnecessary distinction, but the two morphisms are still different.

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 6th 2013

    Toby, I expect Mike knows they are officially morphisms. Cf. #2, where referring to a universal construction just by its underlying representing object is a very standard abuse of language. We all do this on occasion.

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJun 6th 2013

    Thanks for the reminder, though, Toby. I guess I can see that if one assumed that the domain of the kernel of the cokernel would always be isomorphic to the codomain of the cokernel of the kernel, then one might be inclined to define the image to be the kernel of the cokernel, since we do naturally want the image of ff to be a morphism with the same codomain as ff. However, with examples like groups in mind, my inclination would have been to define the image to be the codomain of the cokernel of the kernel equipped with its canonical map to the codomain of ff.

    On the other hand, the topological example does suggest the other choice; thanks Zhen. Does anyone know the history of this terminology?

    • CommentRowNumber12.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013
    • (edited Jun 7th 2013)

    a very standard abuse of language. We all do this on occasion

    Well, sure, but one must drop this abuse of language to understand why the image is one thing rather than the other. So I began by reminding Mike of something that he knew but needed to remember.

    It's also worth remembering that the original (naïve, non-arrow-theoretic) concept of image (in a category of modules, say) is a submodule of the codomain, which is a subobject of the codomain, which is an arrow to the codomain. So even though people might see it as a module, it was always equipped with that arrow rather than the other.

    • CommentRowNumber13.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    with examples like groups in mind, my inclination would have been to define the image to be the codomain of the cokernel of the kernel equipped with its canonical map to the codomain of ff

    Yes, this makes sense. It could never be the cokernel of the kernel itself, since that arrow goes the wrong way, but it certainly could be the arrow suggested here. In an abelian category, this has a simplified description (as the kernel of the cokernel), but the more complicated description is necessary in an arbitrary semiabelian category (such as GrpGrp). I would not be surprised if semiabelian category theorists use ‘image’ in exactly this way.

    • CommentRowNumber14.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    The real definition of image of f:ABf\colon A \to B is relative to a class M\mathbf{M} of morphisms: a universal factorisation of ff as

    AIimfB A \to I \overset{im f}\to B

    with (imf)M(im f) \in \mathbf{M}. Then it becomes a theorem that this has the following descriptions:

    • in an abelian category with M\mathbf{M} the normal monomorphisms, the kernel of the cokernel;
    • in an abelian category with the normal monos, Mike's definition (which is the same thing here);
    • in an abelian category with the regular monos (which are all normal), the kernel of the cokernel;
    • in an abelian category with the regular monos, Mike's definition;
    • in an abelian category with all monos (which are all regular), the kernel of the cokernel;
    • in an abelian category with all monos, Mike's definition;
    • in GrpGrp with the normal monos, the kernel of the cokernel;
    • in GrpGrp with the regular monos (which are not all normal), Mike's definition (which is not the same thing);
    • in GrpGrp with all monos (which are all regular), Mike's definition;
    • in TAGTAG (the category of topological abelian groups) with the normal monos, the kernel of the cokernel;
    • in TAGTAG with the regular monos (which are all normal), the kernel of the cokernel;
    • in TAGTAG with all monos, Mike's definition;
    • in TGTG (the category of topological groups) with the normal monos, the kernel of the cokernel;
    • in TGTG with the regular monos, neither the kernel of the cokernel nor Mike's definition, but something in between;
    • in TGTG with all monos, Mike's definition.

    Each of these categories is concrete, and the set-theoretic image may be recovered using either all monos or the regular monos; where these differ, the subspace topology gives the image using the regular monos. So perhaps that should be the default meaning. This leaves us with the task of describing that in terms of kernels and cokernels in a way that works for TGTG.

    • CommentRowNumber15.
    • CommentAuthorMike Shulman
    • CommentTimeJun 7th 2013

    By the way, the page image often uses the “abuse of language”, denoting by “image” the object rather than the morphism.

    • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 7th 2013

    Toby, just to clarify: my complaint about #9 was that it was (I thought) completely clear what was meant by the question “In what examples do the kernel of the cokernel and the cokernel of the kernel not agree?” Of course they don’t agree as morphisms. But that wasn’t what was being asked.

    But this should be treated as water under the bridge, since everyone is on the same page. Your summary in #14 is useful.

    • CommentRowNumber17.
    • CommentAuthorTobyBartels
    • CommentTimeJun 7th 2013

    Ah, I see. In #9, I was really answering the implied question in

    I’ve never understood why.

    but I didn't quote that part.

    • CommentRowNumber18.
    • CommentAuthorjoe.hannon
    • CommentTimeJun 12th 2013

    I can see that I did kind of move the goalposts on you, Todd. In a question about ordinary singular homology, it’s perfectly reasonable to assume we’re dealing with abelian groups or are in an abelian category. So we should just consider my question answered successfully. Thank you!

    As for whether it holds more generally in an Ab-enriched category, I’m now having doubts. First of all, I guess we’re going to need our morphisms to have kernels and cokernels, and our objects to have biproducts. So maybe the best we can hope for is a pre-Abelian category (finitely complete Ab-category). In fact my guess is that we also need it to be regular (i.e. Abelian), though I’m not sure. In particular, I can’t seem to show that ker(r)Bcoker(i)\ker (r)\to B\to\operatorname{coker}(i) is an isomorphism.

    Actually, this whole proposition seems to be equivalent to the splitting lemma, which I don’t expect to hold in the pre-abelian context. I don’t know a counterexample, but most sources list at least abelian category as a hypothesis. nLab does suggest some stuff about semiabelian categories, but that’s above my pay grade. The upshot is, I’m no longer expecting our statement to hold without some nice assumptions about the category.

    • CommentRowNumber19.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 12th 2013

    The conclusion I arrived at in #4 is that this particular proposition holds in any AbAb-enriched category in which all idempotents split (a rather mild assumption, much milder than finitely complete AbAb-enriched).

    Recall that an idempotent p:XXp: X \to X splits if pp can be written as a composite XrEiXX \stackrel{r}{\to} E \stackrel{i}{\to} X such that ri=1 Er i = 1_E. In this circumstance, ii is the kernel of 1p1 - p and rr is the cokernel of 1p1 - p. We don’t need to assume that kernels or cokernels exist generally – just that kernels and cokernels of idempotents exist (noting that a splitting of an idempotent pp is tantamount to taking a kernel of the complementary idempotent 1p1 - p, or to taking a cokernel of 1p1 - p, as you yourself essentially observed in #3).

    Let me run through the argument. Start from the commutative triangle in #2, where we have r:BAr: B \to A a left inverse to i:ABi: A \to B. Form the idempotent p=irp = i r, and split the complementary idempotent q=1p:BBq = 1-p: B \to B, say as BsEjBB \stackrel{s}{\to} E \stackrel{j}{\to} B. From the above, this means EE is the (underlying object of the) cokernel of pp, as well is also the (underlying object of the) kernel of pp. In this sense, ker(p)Ecoker(p)ker(p) \cong E \cong coker(p).

    Now I claim the kernel of rr exists and is given by E=ker(p)E = ker(p), and also that the cokernel of ii exists and is given by E=coker(p)E = coker(p). In other words,

    ker(r)Ecoker(i),ker(r) \cong E \cong coker(i),

    as was to be shown.

    Proof of claim: the key thing to observe is that r:BAr: B \to A annihilates a morphism f:XBf: X \to B (rf=0r f = 0) iff p:BBp: B \to B annihilates ff. I explained why in #4. In that case, rr annihilates j:EBj: E \to B, since pp does. To see that jj is the kernel of rr, suppose rf=0r f = 0. Then pf=0p f = 0, so f:XBf: X \to B must factor as jhj h for some unique h:XEh: X \to E. We have thus proven that j:EBj: E \to B satisfies the universal property required of ker(r)ker(r). The dual argument shows s:BEs: B \to E is the cokernel of i:ABi: A \to B.

    • CommentRowNumber20.
    • CommentAuthorjoe.hannon
    • CommentTimeJun 13th 2013

    Thank you for explaining again, Todd. Third time’s a charm, it’s perfectly clear now; sorry for being dense. With the axiom that all idempotents are split, the proof goes through. We don’t have all kernels, cokernels, biproducts, and normality of all kernels and cokernels, but we do have these things for splitting of the idempotents, so our proof goes through. Thanks.