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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 $\ker H_0(\epsilon) \cong \operatorname{coker}{H_0(x)}.$ 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 $\ker H_0(\epsilon)$ is a morphism into $H_0(X)$, while $\operatorname{coker}{H_0(x)}$ is a morphism out of $H_0(X).$ 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 $H_0(\epsilon)\circ H_0(x)$ 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

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

where we want to show $coker(f) \cong ker(g)$. 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

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

Let $p = i \circ r: B \to B$; we have $p^2 = p$ since $r \circ i = 1_A$. As subobjects, we can identify $i: A \to B$ with the inclusion $im(p) \hookrightarrow B$; similarly, as quotient objects, we can identify $r: B \to A$ with the projection $B \to im(p)$. Now put $q = 1_B - p$. Then we have $p \circ q = q \circ p = 0$, and $p + q = 1_B$, and also $q^2 = q$. Also note that if $q(b) = 0$, then $b = p(b)$, so $b \in im(p)$.

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

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

which, by the symmetric relationship between $p$ and $q$, is split by the exact sequence

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

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

As you can see, this is an exercise in the yoga of idempotents in $Ab$-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 $\text{im}(p)$. I know three definitions of image of a map $q$, kernel of cokernel of $q$, smallest subobject of codomain through which $q$ 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 $\text{eq}(p,1)=\ker (1-p)=\ker q$ and so $\ker q = \operatorname{coker}q.$ And if $q=js$ with $sj=1$, then $s$ is epi and $j$ is mono, so $\ker q=\ker s$ and $\operatorname{coker}q=\operatorname{coker}j.$ Well we wanted to show that $\ker r=\operatorname{coker}i,$ not $\ker s=\operatorname{coker}j$, but just turn the crank again with those arrows. But I’m still unsure about what category-theoretic properties the objects $\operatorname{im}(p)$ and $\operatorname{im}(q)$ 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 $Ab$-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 $p = i r$ and its complementary idempotent $q = 1 - p$. Split the idempotent $q$. The splitting $q = j s$ can be realized either by taking $j \coloneqq eq(q, 1) = ker(p) = ker(r)$, or $s \coloneqq coeq(q, 1) = coker(p) = coker(i)$. (One should stop for a moment and check the last equalities in those two equation-strings, but the first follows from the observation that $p f = 0$ iff $i r f = 0$ iff $r f = 0$ since $i$ is monic, and similarly the second follows from the fact that $r$ is epic.) Thus we have that the domain of $ker(r)$ is identified with the codomain of $coker(i)$.

• 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 $X \overset{f}\to Y$ factors as

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

People will sometimes replace $I$ 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 $f$ to be a morphism with the same codomain as $f$. 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 $f$.

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 $f$

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 $Grp$). 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 $f\colon A \to B$ is relative to a class $\mathbf{M}$ of morphisms: a universal factorisation of $f$ as

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

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

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

• 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 $\ker (r)\to B\to\operatorname{coker}(i)$ 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 $Ab$-enriched category in which all idempotents split (a rather mild assumption, much milder than finitely complete $Ab$-enriched).

Recall that an idempotent $p: X \to X$ splits if $p$ can be written as a composite $X \stackrel{r}{\to} E \stackrel{i}{\to} X$ such that $r i = 1_E$. In this circumstance, $i$ is the kernel of $1 - p$ and $r$ is the cokernel of $1 - p$. 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 $p$ is tantamount to taking a kernel of the complementary idempotent $1 - p$, or to taking a cokernel of $1 - p$, as you yourself essentially observed in #3).

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

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

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

as was to be shown.

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

• 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.