Proving that $mathrm {Hom}(X,-)$ is left exact in abelian categories
up vote
2
down vote
favorite
I am following Pavel et al's book "Tensor categories".
They claim without proof that the (covariant) functor $F:=mathrm {Hom}(X,-):mathcal C rightarrow textbf{Ab}$ is left exact, where $mathcal C$ is an abelian category. I am trying to show this. I am new to this topic.
Take $0xrightarrow{a} Axrightarrow{b} Bxrightarrow{c} Crightarrow 0$ short exact. We want to show that
$$0rightarrow FArightarrow FBrightarrow FC$$
is exact.
I started by trying to show that im$Fa = ker Fb$. since im$Fa=ker mathrm {coker} Fa$, then $(mathrm {coker}Fa) circ (mathrm{im} Fa)=0$.
I was trying to use this to show that $0xrightarrow{mathrm{im}Fa}FA underset{0}{overset{Fb}{rightrightarrows}} FB$ is an equalizer diagram and thus complete the proof. But I got nowhere.
category-theory abelian-categories
asked Nov 22 at 12:14
Soap
1,020615
add a comment |
up vote
2
down vote
favorite
I am following Pavel et al's book "Tensor categories".
They claim without proof that the (covariant) functor $F:=mathrm {Hom}(X,-):mathcal C rightarrow textbf{Ab}$ is left exact, where $mathcal C$ is an abelian category. I am trying to show this. I am new to this topic.
Take $0xrightarrow{a} Axrightarrow{b} Bxrightarrow{c} Crightarrow 0$ short exact. We want to show that
$$0rightarrow FArightarrow FBrightarrow FC$$
is exact.
I started by trying to show that im$Fa = ker Fb$. since im$Fa=ker mathrm {coker} Fa$, then $(mathrm {coker}Fa) circ (mathrm{im} Fa)=0$.
I was trying to use this to show that $0xrightarrow{mathrm{im}Fa}FA underset{0}{overset{Fb}{rightrightarrows}} FB$ is an equalizer diagram and thus complete the proof. But I got nowhere.
category-theory abelian-categories
asked Nov 22 at 12:14
Soap
1,020615
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am following Pavel et al's book "Tensor categories".
They claim without proof that the (covariant) functor $F:=mathrm {Hom}(X,-):mathcal C rightarrow textbf{Ab}$ is left exact, where $mathcal C$ is an abelian category. I am trying to show this. I am new to this topic.
Take $0xrightarrow{a} Axrightarrow{b} Bxrightarrow{c} Crightarrow 0$ short exact. We want to show that
$$0rightarrow FArightarrow FBrightarrow FC$$
is exact.
I started by trying to show that im$Fa = ker Fb$. since im$Fa=ker mathrm {coker} Fa$, then $(mathrm {coker}Fa) circ (mathrm{im} Fa)=0$.
I was trying to use this to show that $0xrightarrow{mathrm{im}Fa}FA underset{0}{overset{Fb}{rightrightarrows}} FB$ is an equalizer diagram and thus complete the proof. But I got nowhere.
category-theory abelian-categories
asked Nov 22 at 12:14
Soap
1,020615
I am following Pavel et al's book "Tensor categories".
They claim without proof that the (covariant) functor $F:=mathrm {Hom}(X,-):mathcal C rightarrow textbf{Ab}$ is left exact, where $mathcal C$ is an abelian category. I am trying to show this. I am new to this topic.
Take $0xrightarrow{a} Axrightarrow{b} Bxrightarrow{c} Crightarrow 0$ short exact. We want to show that
$$0rightarrow FArightarrow FBrightarrow FC$$
is exact.
I started by trying to show that im$Fa = ker Fb$. since im$Fa=ker mathrm {coker} Fa$, then $(mathrm {coker}Fa) circ (mathrm{im} Fa)=0$.
I was trying to use this to show that $0xrightarrow{mathrm{im}Fa}FA underset{0}{overset{Fb}{rightrightarrows}} FB$ is an equalizer diagram and thus complete the proof. But I got nowhere.
category-theory abelian-categories
category-theory abelian-categories
asked Nov 22 at 12:14
Soap
1,020615
asked Nov 22 at 12:14
Soap
1,020615
asked Nov 22 at 12:14
Soap
1,020615
asked Nov 22 at 12:14
Soap
1,020615
asked Nov 22 at 12:14
Soap
1,020615
1,020615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 to A xrightarrow{iota} B xrightarrow{pi} C to 0$$
be short exact. We want to show that
$$0 to hom(X,A) xrightarrow{iota circ } hom(X,B) xrightarrow{picirc} hom(X,C) $$
is exact. where the morphisms are just the compositions with $iota$ and $pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $iota circ g = iota circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE.
Exactness at $hom(X,A)$:
Here it suffices to show that $iota circ$ is injective, but this just means that $iota circ g =iota circ g'$ implies that $g=g'$, which holds since $iota$ is monic.
Exactness at $hom(X,B)$:
We will prove $mathrm{im}(circiota) subset ker(circ pi)$ and $ker(circ pi) subset mathrm{im}(circiota) $ seperately.
$mathrm{im}(circiota) subset ker(circ pi)$:
Let $f in mathrm{im}(circiota)$ this means that there is a $g$ such that $f=iota circ g$, but this now gives $$pi circ f = pi circ iota circ g = 0 circ g =0$$.
$ker(circ pi) subset mathrm{im}(circiota) $
Let $f in ker(circ pi)$, then we have $pi circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $iota$ actually define a kernel of $pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A xrightarrow{iota} B$, but this literally means that there is a $g$ s.t $f= gcirc iota$ which finishes the claim.
answered Nov 22 at 12:56
Enkidu
1,02618
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
add a comment |
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1 Answer
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Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 to A xrightarrow{iota} B xrightarrow{pi} C to 0$$
be short exact. We want to show that
$$0 to hom(X,A) xrightarrow{iota circ } hom(X,B) xrightarrow{picirc} hom(X,C) $$
is exact. where the morphisms are just the compositions with $iota$ and $pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $iota circ g = iota circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE.
Exactness at $hom(X,A)$:
Here it suffices to show that $iota circ$ is injective, but this just means that $iota circ g =iota circ g'$ implies that $g=g'$, which holds since $iota$ is monic.
Exactness at $hom(X,B)$:
We will prove $mathrm{im}(circiota) subset ker(circ pi)$ and $ker(circ pi) subset mathrm{im}(circiota) $ seperately.
$mathrm{im}(circiota) subset ker(circ pi)$:
Let $f in mathrm{im}(circiota)$ this means that there is a $g$ such that $f=iota circ g$, but this now gives $$pi circ f = pi circ iota circ g = 0 circ g =0$$.
$ker(circ pi) subset mathrm{im}(circiota) $
Let $f in ker(circ pi)$, then we have $pi circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $iota$ actually define a kernel of $pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A xrightarrow{iota} B$, but this literally means that there is a $g$ s.t $f= gcirc iota$ which finishes the claim.
answered Nov 22 at 12:56
Enkidu
1,02618
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
add a comment |
up vote
2
down vote
accepted
Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 to A xrightarrow{iota} B xrightarrow{pi} C to 0$$
be short exact. We want to show that
$$0 to hom(X,A) xrightarrow{iota circ } hom(X,B) xrightarrow{picirc} hom(X,C) $$
is exact. where the morphisms are just the compositions with $iota$ and $pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $iota circ g = iota circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE.
Exactness at $hom(X,A)$:
Here it suffices to show that $iota circ$ is injective, but this just means that $iota circ g =iota circ g'$ implies that $g=g'$, which holds since $iota$ is monic.
Exactness at $hom(X,B)$:
We will prove $mathrm{im}(circiota) subset ker(circ pi)$ and $ker(circ pi) subset mathrm{im}(circiota) $ seperately.
$mathrm{im}(circiota) subset ker(circ pi)$:
Let $f in mathrm{im}(circiota)$ this means that there is a $g$ such that $f=iota circ g$, but this now gives $$pi circ f = pi circ iota circ g = 0 circ g =0$$.
$ker(circ pi) subset mathrm{im}(circiota) $
Let $f in ker(circ pi)$, then we have $pi circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $iota$ actually define a kernel of $pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A xrightarrow{iota} B$, but this literally means that there is a $g$ s.t $f= gcirc iota$ which finishes the claim.
answered Nov 22 at 12:56
Enkidu
1,02618
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 to A xrightarrow{iota} B xrightarrow{pi} C to 0$$
be short exact. We want to show that
$$0 to hom(X,A) xrightarrow{iota circ } hom(X,B) xrightarrow{picirc} hom(X,C) $$
is exact. where the morphisms are just the compositions with $iota$ and $pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $iota circ g = iota circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE.
Exactness at $hom(X,A)$:
Here it suffices to show that $iota circ$ is injective, but this just means that $iota circ g =iota circ g'$ implies that $g=g'$, which holds since $iota$ is monic.
Exactness at $hom(X,B)$:
We will prove $mathrm{im}(circiota) subset ker(circ pi)$ and $ker(circ pi) subset mathrm{im}(circiota) $ seperately.
$mathrm{im}(circiota) subset ker(circ pi)$:
Let $f in mathrm{im}(circiota)$ this means that there is a $g$ such that $f=iota circ g$, but this now gives $$pi circ f = pi circ iota circ g = 0 circ g =0$$.
$ker(circ pi) subset mathrm{im}(circiota) $
Let $f in ker(circ pi)$, then we have $pi circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $iota$ actually define a kernel of $pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A xrightarrow{iota} B$, but this literally means that there is a $g$ s.t $f= gcirc iota$ which finishes the claim.
answered Nov 22 at 12:56
Enkidu
1,02618
Don't worry, i know it is tricky in the beginning, however, May i use some different notation?
I.e.: Let $$0 to A xrightarrow{iota} B xrightarrow{pi} C to 0$$
be short exact. We want to show that
$$0 to hom(X,A) xrightarrow{iota circ } hom(X,B) xrightarrow{picirc} hom(X,C) $$
is exact. where the morphisms are just the compositions with $iota$ and $pi$.
Now first remember that injective maps are monomorphisms (I call them monic), i.e. $iota circ g = iota circ g'$ implies $g=g'$ for any $g,g'$, if you do not know this, it is an awesome EXERCISE.
Exactness at $hom(X,A)$:
Here it suffices to show that $iota circ$ is injective, but this just means that $iota circ g =iota circ g'$ implies that $g=g'$, which holds since $iota$ is monic.
Exactness at $hom(X,B)$:
We will prove $mathrm{im}(circiota) subset ker(circ pi)$ and $ker(circ pi) subset mathrm{im}(circiota) $ seperately.
$mathrm{im}(circiota) subset ker(circ pi)$:
Let $f in mathrm{im}(circiota)$ this means that there is a $g$ such that $f=iota circ g$, but this now gives $$pi circ f = pi circ iota circ g = 0 circ g =0$$.
$ker(circ pi) subset mathrm{im}(circiota) $
Let $f in ker(circ pi)$, then we have $pi circ f =0$, but this means by the universal property of the kernel that $f$ factors over the kernel. Now since $A$ and $iota$ actually define a kernel of $pi$ (kernels are only unique up to unique isomorphism) $f$ factors over $A xrightarrow{iota} B$, but this literally means that there is a $g$ s.t $f= gcirc iota$ which finishes the claim.
answered Nov 22 at 12:56
Enkidu
1,02618
answered Nov 22 at 12:56
Enkidu
1,02618
answered Nov 22 at 12:56
Enkidu
1,02618
answered Nov 22 at 12:56
Enkidu
1,02618
1,02618
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
add a comment |
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
Very good answer. But I have a couple of questions: 1. You say that $iota$ is a kernel of $pi$, but it is im $iota$ instead, no? 2. You use that $ker c = mathrm{im} b Rightarrow cb=0$. But is this valid in general? I tried to show it using the definition of im $b = ker mathrm{coker} b$ but didn't get it.
– Soap
Nov 23 at 15:52
1
1
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
No, $(mathrm{im}(iota), mathrm{inclusion})$ is what you know as a kernel, however, in a pointed category a kernel is actually defined over a universal property and consists of an object and a morphism. Hence, here I literally mean that $(A,iota)$ is a kernel! Yea, it is valid in general, but it should not need the natural isomorphism between the image and the coimage, but more the uniqueness of the zero morphism (this holds in any pointed category), this should be a direct corollary of the universal property of the kernel.
– Enkidu
Nov 26 at 8:53
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
I understand that I can write $b=im(b) circ e$, and since $cb=0$ then this $e$ is unique. But it is not necesserily an isomorphism, right? So I still don't see why $b$ is a kernel...
– Soap
Dec 1 at 14:14
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
Well that is precisely the property to be a short exact sequence in an abelian category. There are different equivalent conditions you need additionally to the existence of kernels and cokernels. Something like either: "coimage and image are naturally isomorphic" or "monics are kernels of their cokernels and epics are cokernels of their kernels" something like that. And that is precisely what comes into play here
– Enkidu
Dec 3 at 9:57
add a comment |
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