Proving the snake lemma without a diagram chase
up vote
40
down vote
favorite
Suppose we have two short exact sequences in an abelian category
$$0 to A mathrel{overset{f}{to}} B mathrel{overset{g}{to}} C to 0 $$
$$0 to A' mathrel{overset{f'}{to}} B' mathrel{overset{g'}{to}} C' to 0 $$
and morphisms $a : A to A', b : B to B', c : C to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence
$$0 to ker a to ker b to ker c to operatorname{coker} a to operatorname{coker} b to operatorname{coker} c to 0$$
where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.
It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $ker a, ker b, operatorname{coker} b, operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : ker c to operatorname{coker} a$. However, I'm stuck showing exactness at $ker c$ and $operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
commutative-algebra category-theory homological-algebra abelian-categories
add a comment |
up vote
40
down vote
favorite
Suppose we have two short exact sequences in an abelian category
$$0 to A mathrel{overset{f}{to}} B mathrel{overset{g}{to}} C to 0 $$
$$0 to A' mathrel{overset{f'}{to}} B' mathrel{overset{g'}{to}} C' to 0 $$
and morphisms $a : A to A', b : B to B', c : C to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence
$$0 to ker a to ker b to ker c to operatorname{coker} a to operatorname{coker} b to operatorname{coker} c to 0$$
where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.
It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $ker a, ker b, operatorname{coker} b, operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : ker c to operatorname{coker} a$. However, I'm stuck showing exactness at $ker c$ and $operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
commutative-algebra category-theory homological-algebra abelian-categories
4
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
10
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
4
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18
add a comment |
up vote
40
down vote
favorite
up vote
40
down vote
favorite
Suppose we have two short exact sequences in an abelian category
$$0 to A mathrel{overset{f}{to}} B mathrel{overset{g}{to}} C to 0 $$
$$0 to A' mathrel{overset{f'}{to}} B' mathrel{overset{g'}{to}} C' to 0 $$
and morphisms $a : A to A', b : B to B', c : C to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence
$$0 to ker a to ker b to ker c to operatorname{coker} a to operatorname{coker} b to operatorname{coker} c to 0$$
where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.
It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $ker a, ker b, operatorname{coker} b, operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : ker c to operatorname{coker} a$. However, I'm stuck showing exactness at $ker c$ and $operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
commutative-algebra category-theory homological-algebra abelian-categories
Suppose we have two short exact sequences in an abelian category
$$0 to A mathrel{overset{f}{to}} B mathrel{overset{g}{to}} C to 0 $$
$$0 to A' mathrel{overset{f'}{to}} B' mathrel{overset{g'}{to}} C' to 0 $$
and morphisms $a : A to A', b : B to B', c : C to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence
$$0 to ker a to ker b to ker c to operatorname{coker} a to operatorname{coker} b to operatorname{coker} c to 0$$
where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.
It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $ker a, ker b, operatorname{coker} b, operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : ker c to operatorname{coker} a$. However, I'm stuck showing exactness at $ker c$ and $operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
commutative-algebra category-theory homological-algebra abelian-categories
commutative-algebra category-theory homological-algebra abelian-categories
asked Oct 22 '11 at 16:47
Zhen Lin
59.9k4104215
59.9k4104215
4
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
10
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
4
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18
add a comment |
4
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
10
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
4
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18
4
4
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
10
10
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
4
4
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18
add a comment |
4 Answers
4
active
oldest
votes
up vote
27
down vote
accepted
You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.
In any case, you can also construct the boundary map as follows:
We are given a map $b: B to B'$. Let $B'' hookrightarrow B$ denote
the preimage in $B$ of $ker c$. (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B to C to C'$.)
Then the map $B''hookrightarrow B rightarrow B'$ factors through the monomorphism $A' hookrightarrow B'$
(using the fact that $A' =ker(B' to C'), , $).
This then induces a map on quotients $ B''/A to A'/operatorname{im}A$,
which is precisely the desired map $ker c tooperatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
add a comment |
up vote
9
down vote
There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
add a comment |
up vote
5
down vote
The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says:
If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
add a comment |
up vote
3
down vote
There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
27
down vote
accepted
You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.
In any case, you can also construct the boundary map as follows:
We are given a map $b: B to B'$. Let $B'' hookrightarrow B$ denote
the preimage in $B$ of $ker c$. (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B to C to C'$.)
Then the map $B''hookrightarrow B rightarrow B'$ factors through the monomorphism $A' hookrightarrow B'$
(using the fact that $A' =ker(B' to C'), , $).
This then induces a map on quotients $ B''/A to A'/operatorname{im}A$,
which is precisely the desired map $ker c tooperatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
add a comment |
up vote
27
down vote
accepted
You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.
In any case, you can also construct the boundary map as follows:
We are given a map $b: B to B'$. Let $B'' hookrightarrow B$ denote
the preimage in $B$ of $ker c$. (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B to C to C'$.)
Then the map $B''hookrightarrow B rightarrow B'$ factors through the monomorphism $A' hookrightarrow B'$
(using the fact that $A' =ker(B' to C'), , $).
This then induces a map on quotients $ B''/A to A'/operatorname{im}A$,
which is precisely the desired map $ker c tooperatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
add a comment |
up vote
27
down vote
accepted
up vote
27
down vote
accepted
You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.
In any case, you can also construct the boundary map as follows:
We are given a map $b: B to B'$. Let $B'' hookrightarrow B$ denote
the preimage in $B$ of $ker c$. (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B to C to C'$.)
Then the map $B''hookrightarrow B rightarrow B'$ factors through the monomorphism $A' hookrightarrow B'$
(using the fact that $A' =ker(B' to C'), , $).
This then induces a map on quotients $ B''/A to A'/operatorname{im}A$,
which is precisely the desired map $ker c tooperatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.
You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.
In any case, you can also construct the boundary map as follows:
We are given a map $b: B to B'$. Let $B'' hookrightarrow B$ denote
the preimage in $B$ of $ker c$. (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B to C to C'$.)
Then the map $B''hookrightarrow B rightarrow B'$ factors through the monomorphism $A' hookrightarrow B'$
(using the fact that $A' =ker(B' to C'), , $).
This then induces a map on quotients $ B''/A to A'/operatorname{im}A$,
which is precisely the desired map $ker c tooperatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.
edited Jan 19 '12 at 8:10
answered Oct 22 '11 at 16:56
Matt E
104k8215381
104k8215381
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
add a comment |
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
1
1
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
Your construction of the connecting morphism is very concise, thanks. I still have to work out how to prove exactness though.
– Zhen Lin
Oct 22 '11 at 20:31
1
1
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
@ZhenLin I've been trying to figure out how to actually get the connecting morphism (induced map on quotients) but I haven't managed to. Can you help me?
– user153312
Jan 20 '15 at 19:08
1
1
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
There are many ways to do it. Another way is to use the mapping cone construction.
– Zhen Lin
Jan 20 '15 at 21:17
add a comment |
up vote
9
down vote
There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
add a comment |
up vote
9
down vote
There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
add a comment |
up vote
9
down vote
up vote
9
down vote
There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
answered Jan 14 '13 at 7:30
Brian Fitzpatrick
21k42958
21k42958
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
add a comment |
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
18
18
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
Can you provide any details or at least some ideas of the proof here, for those of us who don't have the cited book?
– robjohn♦
Jan 14 '13 at 7:37
add a comment |
up vote
5
down vote
The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says:
If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
add a comment |
up vote
5
down vote
The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says:
If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
add a comment |
up vote
5
down vote
up vote
5
down vote
The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says:
If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.
The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says:
If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.
edited Nov 23 at 5:51
user26857
39.2k123882
39.2k123882
answered Mar 10 '13 at 2:39
Mozibur Ullah
2,6651029
2,6651029
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
add a comment |
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
By the way, did you read Bergman's early paper about diagram chasing in general Abelian categories? I don't understand the part of addition: how did he add two generalized elements $fcolon Ato X$ and $gcolon Bto X$?
– Frank Science
Oct 7 '16 at 16:10
add a comment |
up vote
3
down vote
There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
add a comment |
up vote
3
down vote
There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
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There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
edited Oct 22 '11 at 20:05
t.b.
61.9k7203285
61.9k7203285
answered Oct 22 '11 at 20:01
Rafael Mrđen
944823
944823
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
add a comment |
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
1
1
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
I think that's essentially the same as the one Matt E's mentioning at the beginning of his answer and is carried out in MacLane's book.
– t.b.
Oct 22 '11 at 20:12
add a comment |
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4
In the comments to your question on references for sheaf cohomology, Zev and I gave you two links that contain element-free proofs: 1. Lambek's review of Strooker's book 2. and my survey on Exact categories where a detailed proof is given (for exact categories) in section 8.
– t.b.
Oct 22 '11 at 17:09
10
You can also watch the opening minutes of It's My Turn with Jill Clayburgh to see a proof...
– Arturo Magidin
Oct 22 '11 at 19:05
4
@Arturo: It's on youtube of course. However, it's an "elementary" proof.
– t.b.
Oct 22 '11 at 19:18