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Exact sequence involving pushout

Counterexample for a pullback-pushout situationPushout on the following diagramOn pushouts and mapping cylinders in exact categoriesConcepts like a pushout or pullback but slightly differentIs there a name for taking the pushforward (ie pushout) over the pullback?Infinite product of a short exact sequenceDefinition of a pushout of a short exact sequenceEvery abelian category is exact categoryFactorize a morphism into a short exact sequenceone question about homotopy pushout

1

$begingroup$

I think I have a bit of a misunderstanding of pushouts. Let $U, X, Y$ be objects of an abelian category $A$ with morphisms $f: U rightarrow X$ and $g: U rightarrow Y.$ Let $Z$ be the pushout of this diagram with natural morphisms $f': X rightarrow Z$ and $g': Y rightarrow Z.$ Then apparently we have a short exact sequence $$0 rightarrow U rightarrow X oplus Y rightarrow Z rightarrow 0 $$ where the map $U rightarrow X oplus Y$ is defined as $u mapsto (f(u), g(u))$ and the map $X oplus Y rightarrow Z$ is denoted by $(f', g').$ I cannot see how this is true. My book states the mapping of the short exact sequence in this way but I think the map $X oplus Y rightarrow Z$ should rather be $(-f', g')$ or $(f', -g')$ where $(-f', g')(x, y) = -f'(x) + g'(y).$ Am I right or am I misunderstanding something?

Here is a screenshot of the passage. The arrow notation is a bit different...

category-theory

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asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

$endgroup$

add a comment | 

1

$begingroup$

I think I have a bit of a misunderstanding of pushouts. Let $U, X, Y$ be objects of an abelian category $A$ with morphisms $f: U rightarrow X$ and $g: U rightarrow Y.$ Let $Z$ be the pushout of this diagram with natural morphisms $f': X rightarrow Z$ and $g': Y rightarrow Z.$ Then apparently we have a short exact sequence $$0 rightarrow U rightarrow X oplus Y rightarrow Z rightarrow 0 $$ where the map $U rightarrow X oplus Y$ is defined as $u mapsto (f(u), g(u))$ and the map $X oplus Y rightarrow Z$ is denoted by $(f', g').$ I cannot see how this is true. My book states the mapping of the short exact sequence in this way but I think the map $X oplus Y rightarrow Z$ should rather be $(-f', g')$ or $(f', -g')$ where $(-f', g')(x, y) = -f'(x) + g'(y).$ Am I right or am I misunderstanding something?

Here is a screenshot of the passage. The arrow notation is a bit different...

category-theory

share|cite|improve this question

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

$endgroup$

add a comment | 

$begingroup$

I think I have a bit of a misunderstanding of pushouts. Let $U, X, Y$ be objects of an abelian category $A$ with morphisms $f: U rightarrow X$ and $g: U rightarrow Y.$ Let $Z$ be the pushout of this diagram with natural morphisms $f': X rightarrow Z$ and $g': Y rightarrow Z.$ Then apparently we have a short exact sequence $$0 rightarrow U rightarrow X oplus Y rightarrow Z rightarrow 0 $$ where the map $U rightarrow X oplus Y$ is defined as $u mapsto (f(u), g(u))$ and the map $X oplus Y rightarrow Z$ is denoted by $(f', g').$ I cannot see how this is true. My book states the mapping of the short exact sequence in this way but I think the map $X oplus Y rightarrow Z$ should rather be $(-f', g')$ or $(f', -g')$ where $(-f', g')(x, y) = -f'(x) + g'(y).$ Am I right or am I misunderstanding something?

Here is a screenshot of the passage. The arrow notation is a bit different...

category-theory

share|cite|improve this question

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

$endgroup$

share|cite|improve this question

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

share|cite|improve this question

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

asked Mar 11 at 8:45

伽罗瓦伽罗瓦

1,219615

1 Answer
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$begingroup$

$requireAMScd$In an abelian category, a commutative square
beginCD
U@>f>> X\
@VgVV@VVg'V\
Y@>>f'> Z
endCD
is a pushout if and only if the sequence
$$Uxrightarrow[f,g]Xoplus Yxrightarrowleft[genfrac0ptg'-f'right]Zto 0$$
is exact.

The exactness of the sequence is equivalent to the fact that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]left[genfrac0ptg'-f'right]=fg'-gf'=0$$
On the other hand, if $[f,g]left[genfrac0ptpqright]=fp+gq=0:Uto V$, then $fp=-gq$ hence there exists one and only one morphism $h:Zto V$ such that $p=g'h$ and $-q=f'h$, hence
beginalign
left[genfrac0ptpqright]
&=left[genfrac0ptg'h-f'hright]\
&=left[genfrac0ptg'-f'right]h
endalign
thus proving that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.

share|cite|improve this answer

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

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add a comment | 

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0

$begingroup$

$requireAMScd$In an abelian category, a commutative square
beginCD
U@>f>> X\
@VgVV@VVg'V\
Y@>>f'> Z
endCD
is a pushout if and only if the sequence
$$Uxrightarrow[f,g]Xoplus Yxrightarrowleft[genfrac0ptg'-f'right]Zto 0$$
is exact.

The exactness of the sequence is equivalent to the fact that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]left[genfrac0ptg'-f'right]=fg'-gf'=0$$
On the other hand, if $[f,g]left[genfrac0ptpqright]=fp+gq=0:Uto V$, then $fp=-gq$ hence there exists one and only one morphism $h:Zto V$ such that $p=g'h$ and $-q=f'h$, hence
beginalign
left[genfrac0ptpqright]
&=left[genfrac0ptg'h-f'hright]\
&=left[genfrac0ptg'-f'right]h
endalign
thus proving that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.

share|cite|improve this answer

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

$endgroup$

add a comment | 

0

$begingroup$

$requireAMScd$In an abelian category, a commutative square
beginCD
U@>f>> X\
@VgVV@VVg'V\
Y@>>f'> Z
endCD
is a pushout if and only if the sequence
$$Uxrightarrow[f,g]Xoplus Yxrightarrowleft[genfrac0ptg'-f'right]Zto 0$$
is exact.

The exactness of the sequence is equivalent to the fact that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]left[genfrac0ptg'-f'right]=fg'-gf'=0$$
On the other hand, if $[f,g]left[genfrac0ptpqright]=fp+gq=0:Uto V$, then $fp=-gq$ hence there exists one and only one morphism $h:Zto V$ such that $p=g'h$ and $-q=f'h$, hence
beginalign
left[genfrac0ptpqright]
&=left[genfrac0ptg'h-f'hright]\
&=left[genfrac0ptg'-f'right]h
endalign
thus proving that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.

share|cite|improve this answer

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

$endgroup$

add a comment | 

$begingroup$

$requireAMScd$In an abelian category, a commutative square
beginCD
U@>f>> X\
@VgVV@VVg'V\
Y@>>f'> Z
endCD
is a pushout if and only if the sequence
$$Uxrightarrow[f,g]Xoplus Yxrightarrowleft[genfrac0ptg'-f'right]Zto 0$$
is exact.

The exactness of the sequence is equivalent to the fact that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]left[genfrac0ptg'-f'right]=fg'-gf'=0$$
On the other hand, if $[f,g]left[genfrac0ptpqright]=fp+gq=0:Uto V$, then $fp=-gq$ hence there exists one and only one morphism $h:Zto V$ such that $p=g'h$ and $-q=f'h$, hence
beginalign
left[genfrac0ptpqright]
&=left[genfrac0ptg'h-f'hright]\
&=left[genfrac0ptg'-f'right]h
endalign
thus proving that $left[genfrac0ptg'-f'right]$ is the cokernel of $[f,g]$.

share|cite|improve this answer

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426

answered Mar 11 at 11:06

Fabio LucchiniFabio Lucchini

8,88311426