Mixing
Is additivity necessary for a left exact functor to preserve pullbacks?
$begingroup$
I'm having a bit of difficulty with exercise 5.16 from Rotman's An Introduction to Homological Algebra (second edition).
The exercise (at least the relevant part) reads
Prove that every left exact covariant functor $T:{}_Rmathbf{Mod} to mathbf{Ab}$ preserves pullbacks.
Now, I went back and checked the definition, and a covariant functor is defined to be left exact if
$$ 0to Ato Bto C$$ being exact implies
$$ 0to TA to TBto TC$$
is exact, with no mention of additivity.
Now, I'm aware of this question, but the comment on the question doesn't apply given the definition in Rotman, and the accepted answer to the question doesn't get me any further than the work I've already done. Thus this new question.
My work, and a precise statement of exactly where I'm running into trouble without additivity
It's not hard to show that left exact functors preserve direct sums,
and it's also not hard to check that
$$require{AMScd}begin{CD}
D @>>> C
\
@VVV @VVgV
\
B @>>f> A
end{CD}$$
is a pullback square if and only if
$$0 to D to Boplus C newcommandtobyxrightarrow toby{fpi_B-gpi_C} A $$
is exact.
Then if we apply $T$ to this exact sequence, we get the exact sequence
$$ 0 to TD to TB oplus TC toby{T(fpi_B-gpi_C)} TA. $$
Now when $T$ is additive, we have
$$T(fpi_B -gpi_C) = (Tf)pi_{TB}-(Tg)pi_{TC},$$
so this exact sequence is the correct exact sequence to show that
$$begin{CD}
TD @>>> TC
\
@VVV @VVTgV
\
TB @>>Tf> TA
end{CD}
$$
is a pullback square.
However if $T$ is not additive, then it is not clear that the exact sequence we get from applying $T$ to the exact sequence from the pullback square in ${}_Rmathbf{Mod}$ is the correct exact sequence to prove that $TD$ is the pullback in $mathbf{Ab}$. This is where I am stuck.
The question
Hence my questions:
- Is there a way to finish the proof without using additivity?
- Is there something special about these particular categories?
- Is there perhaps a counterexample, showing that we need to assume additivity? (I'm not aware of any left exact but not additive functors (nor is Google, based on a quick search), so I don't even know where to begin searching for a counterexample)
homological-algebra exact-sequence abelian-categories functors
asked Jan 7 at 18:47
jgonjgon
13.7k22041
$endgroup$
add a comment |
$begingroup$
I'm having a bit of difficulty with exercise 5.16 from Rotman's An Introduction to Homological Algebra (second edition).
The exercise (at least the relevant part) reads
Prove that every left exact covariant functor $T:{}_Rmathbf{Mod} to mathbf{Ab}$ preserves pullbacks.
Now, I went back and checked the definition, and a covariant functor is defined to be left exact if
$$ 0to Ato Bto C$$ being exact implies
$$ 0to TA to TBto TC$$
is exact, with no mention of additivity.
Now, I'm aware of this question, but the comment on the question doesn't apply given the definition in Rotman, and the accepted answer to the question doesn't get me any further than the work I've already done. Thus this new question.
My work, and a precise statement of exactly where I'm running into trouble without additivity
It's not hard to show that left exact functors preserve direct sums,
and it's also not hard to check that
$$require{AMScd}begin{CD}
D @>>> C
\
@VVV @VVgV
\
B @>>f> A
end{CD}$$
is a pullback square if and only if
$$0 to D to Boplus C newcommandtobyxrightarrow toby{fpi_B-gpi_C} A $$
is exact.
Then if we apply $T$ to this exact sequence, we get the exact sequence
$$ 0 to TD to TB oplus TC toby{T(fpi_B-gpi_C)} TA. $$
Now when $T$ is additive, we have
$$T(fpi_B -gpi_C) = (Tf)pi_{TB}-(Tg)pi_{TC},$$
so this exact sequence is the correct exact sequence to show that
$$begin{CD}
TD @>>> TC
\
@VVV @VVTgV
\
TB @>>Tf> TA
end{CD}
$$
is a pullback square.
However if $T$ is not additive, then it is not clear that the exact sequence we get from applying $T$ to the exact sequence from the pullback square in ${}_Rmathbf{Mod}$ is the correct exact sequence to prove that $TD$ is the pullback in $mathbf{Ab}$. This is where I am stuck.
The question
Hence my questions:
- Is there a way to finish the proof without using additivity?
- Is there something special about these particular categories?
- Is there perhaps a counterexample, showing that we need to assume additivity? (I'm not aware of any left exact but not additive functors (nor is Google, based on a quick search), so I don't even know where to begin searching for a counterexample)
homological-algebra exact-sequence abelian-categories functors
asked Jan 7 at 18:47
jgonjgon
13.7k22041
$endgroup$
1
$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10
add a comment |
$begingroup$
I'm having a bit of difficulty with exercise 5.16 from Rotman's An Introduction to Homological Algebra (second edition).
The exercise (at least the relevant part) reads
Prove that every left exact covariant functor $T:{}_Rmathbf{Mod} to mathbf{Ab}$ preserves pullbacks.
Now, I went back and checked the definition, and a covariant functor is defined to be left exact if
$$ 0to Ato Bto C$$ being exact implies
$$ 0to TA to TBto TC$$
is exact, with no mention of additivity.
Now, I'm aware of this question, but the comment on the question doesn't apply given the definition in Rotman, and the accepted answer to the question doesn't get me any further than the work I've already done. Thus this new question.
My work, and a precise statement of exactly where I'm running into trouble without additivity
It's not hard to show that left exact functors preserve direct sums,
and it's also not hard to check that
$$require{AMScd}begin{CD}
D @>>> C
\
@VVV @VVgV
\
B @>>f> A
end{CD}$$
is a pullback square if and only if
$$0 to D to Boplus C newcommandtobyxrightarrow toby{fpi_B-gpi_C} A $$
is exact.
Then if we apply $T$ to this exact sequence, we get the exact sequence
$$ 0 to TD to TB oplus TC toby{T(fpi_B-gpi_C)} TA. $$
Now when $T$ is additive, we have
$$T(fpi_B -gpi_C) = (Tf)pi_{TB}-(Tg)pi_{TC},$$
so this exact sequence is the correct exact sequence to show that
$$begin{CD}
TD @>>> TC
\
@VVV @VVTgV
\
TB @>>Tf> TA
end{CD}
$$
is a pullback square.
However if $T$ is not additive, then it is not clear that the exact sequence we get from applying $T$ to the exact sequence from the pullback square in ${}_Rmathbf{Mod}$ is the correct exact sequence to prove that $TD$ is the pullback in $mathbf{Ab}$. This is where I am stuck.
The question
Hence my questions:
- Is there a way to finish the proof without using additivity?
- Is there something special about these particular categories?
- Is there perhaps a counterexample, showing that we need to assume additivity? (I'm not aware of any left exact but not additive functors (nor is Google, based on a quick search), so I don't even know where to begin searching for a counterexample)
homological-algebra exact-sequence abelian-categories functors
asked Jan 7 at 18:47
jgonjgon
13.7k22041
$endgroup$
I'm having a bit of difficulty with exercise 5.16 from Rotman's An Introduction to Homological Algebra (second edition).
The exercise (at least the relevant part) reads
Prove that every left exact covariant functor $T:{}_Rmathbf{Mod} to mathbf{Ab}$ preserves pullbacks.
Now, I went back and checked the definition, and a covariant functor is defined to be left exact if
$$ 0to Ato Bto C$$ being exact implies
$$ 0to TA to TBto TC$$
is exact, with no mention of additivity.
Now, I'm aware of this question, but the comment on the question doesn't apply given the definition in Rotman, and the accepted answer to the question doesn't get me any further than the work I've already done. Thus this new question.
My work, and a precise statement of exactly where I'm running into trouble without additivity
It's not hard to show that left exact functors preserve direct sums,
and it's also not hard to check that
$$require{AMScd}begin{CD}
D @>>> C
\
@VVV @VVgV
\
B @>>f> A
end{CD}$$
is a pullback square if and only if
$$0 to D to Boplus C newcommandtobyxrightarrow toby{fpi_B-gpi_C} A $$
is exact.
Then if we apply $T$ to this exact sequence, we get the exact sequence
$$ 0 to TD to TB oplus TC toby{T(fpi_B-gpi_C)} TA. $$
Now when $T$ is additive, we have
$$T(fpi_B -gpi_C) = (Tf)pi_{TB}-(Tg)pi_{TC},$$
so this exact sequence is the correct exact sequence to show that
$$begin{CD}
TD @>>> TC
\
@VVV @VVTgV
\
TB @>>Tf> TA
end{CD}
$$
is a pullback square.
However if $T$ is not additive, then it is not clear that the exact sequence we get from applying $T$ to the exact sequence from the pullback square in ${}_Rmathbf{Mod}$ is the correct exact sequence to prove that $TD$ is the pullback in $mathbf{Ab}$. This is where I am stuck.
The question
Hence my questions:
- Is there a way to finish the proof without using additivity?
- Is there something special about these particular categories?
- Is there perhaps a counterexample, showing that we need to assume additivity? (I'm not aware of any left exact but not additive functors (nor is Google, based on a quick search), so I don't even know where to begin searching for a counterexample)
homological-algebra exact-sequence abelian-categories functors
homological-algebra exact-sequence abelian-categories functors
asked Jan 7 at 18:47
jgonjgon
13.7k22041
asked Jan 7 at 18:47
jgonjgon
13.7k22041
asked Jan 7 at 18:47
jgonjgon
13.7k22041
asked Jan 7 at 18:47
jgonjgon
13.7k22041
asked Jan 7 at 18:47
jgonjgon
13.7k22041
13.7k22041
1
$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10
add a comment |
1
$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10
1
1
$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : Ato B$ are morphisms, we define $f+g$ to be the composite
$$ Anewcommandtobyxrightarrowtoby{Delta_A} A oplus Atoby{foplus g}Boplus B toby{nabla_B} B,$$
where $Delta_A : Ato Aoplus A$ is the diagonal map and $nabla_B : Boplus Bto B$ is the codiagonal.
Then clearly any functor which satisfies $T(foplus g)=Tfoplus Tg$ and $TDelta_A = Delta_{TA}$ and $Tnabla_B = nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(Aoplus B)=TAoplus TB$ and $Tiota_X=iota_{TX}$ and $Tpi_X = pi_{TX}$ for the structure morphisms $iota_A,iota_B,pi_A,$ and $pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $Delta_A$ is the map such that $pi_{A,i}Delta_A=newcommandid{operatorname{id}}id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $TDelta_A$ satisfies $pi_{TA,i}T(Delta_A)=id_{TA}$ for $i=1,2$, thus $TDelta_A=Delta_{TA}$. A dual argument shows that $Tnabla_B=nabla_{TB}$, and finally $foplus g$ is defined to be the map such that
$$pi_{B,1}(foplus g)iota_{A,1} = f,$$
$$pi_{B,2}(foplus g)iota_{A,2} = g,$$
$$pi_{B,1}(foplus g)iota_{A,2} = 0,$$
and
$$pi_{B,2}(foplus g)iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(foplus g)$ satisfies the requirement to be $Tfoplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TAto TB to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 to 0 toby{id_0} 0 toby{id_0} 0.$$
This is clearly exact,
so
$$ 0 to T0 toby{id_{T0}} T0 toby{id_{T0}} T0$$
is exact, which implies that $$0=ker id_{T0} = T0=operatorname{im}_{id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $pi_Aiota_A =id_A$, so $(Tpi_A)(Tiota_A)=id_{TA}$. This also applies to $pi_Biota_B$, so $Tpi_B$ must be an epimorphism. Hence, exactness of
$$0to A toby{iota_A} Aoplus B toby{pi_B} B to 0$$
not only implies exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB, $$
but actually the exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB to 0,$$
and the fact that $(Tpi_A)(Tiota_A)=id_{TA}$ implies that this exact sequence is split, giving
$T(Aoplus B) simeq TAoplus TB$.
answered Jan 7 at 22:01
jgonjgon
13.7k22041
$endgroup$
add a comment |
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$begingroup$
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : Ato B$ are morphisms, we define $f+g$ to be the composite
$$ Anewcommandtobyxrightarrowtoby{Delta_A} A oplus Atoby{foplus g}Boplus B toby{nabla_B} B,$$
where $Delta_A : Ato Aoplus A$ is the diagonal map and $nabla_B : Boplus Bto B$ is the codiagonal.
Then clearly any functor which satisfies $T(foplus g)=Tfoplus Tg$ and $TDelta_A = Delta_{TA}$ and $Tnabla_B = nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(Aoplus B)=TAoplus TB$ and $Tiota_X=iota_{TX}$ and $Tpi_X = pi_{TX}$ for the structure morphisms $iota_A,iota_B,pi_A,$ and $pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $Delta_A$ is the map such that $pi_{A,i}Delta_A=newcommandid{operatorname{id}}id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $TDelta_A$ satisfies $pi_{TA,i}T(Delta_A)=id_{TA}$ for $i=1,2$, thus $TDelta_A=Delta_{TA}$. A dual argument shows that $Tnabla_B=nabla_{TB}$, and finally $foplus g$ is defined to be the map such that
$$pi_{B,1}(foplus g)iota_{A,1} = f,$$
$$pi_{B,2}(foplus g)iota_{A,2} = g,$$
$$pi_{B,1}(foplus g)iota_{A,2} = 0,$$
and
$$pi_{B,2}(foplus g)iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(foplus g)$ satisfies the requirement to be $Tfoplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TAto TB to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 to 0 toby{id_0} 0 toby{id_0} 0.$$
This is clearly exact,
so
$$ 0 to T0 toby{id_{T0}} T0 toby{id_{T0}} T0$$
is exact, which implies that $$0=ker id_{T0} = T0=operatorname{im}_{id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $pi_Aiota_A =id_A$, so $(Tpi_A)(Tiota_A)=id_{TA}$. This also applies to $pi_Biota_B$, so $Tpi_B$ must be an epimorphism. Hence, exactness of
$$0to A toby{iota_A} Aoplus B toby{pi_B} B to 0$$
not only implies exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB, $$
but actually the exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB to 0,$$
and the fact that $(Tpi_A)(Tiota_A)=id_{TA}$ implies that this exact sequence is split, giving
$T(Aoplus B) simeq TAoplus TB$.
answered Jan 7 at 22:01
jgonjgon
13.7k22041
$endgroup$
add a comment |
$begingroup$
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : Ato B$ are morphisms, we define $f+g$ to be the composite
$$ Anewcommandtobyxrightarrowtoby{Delta_A} A oplus Atoby{foplus g}Boplus B toby{nabla_B} B,$$
where $Delta_A : Ato Aoplus A$ is the diagonal map and $nabla_B : Boplus Bto B$ is the codiagonal.
Then clearly any functor which satisfies $T(foplus g)=Tfoplus Tg$ and $TDelta_A = Delta_{TA}$ and $Tnabla_B = nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(Aoplus B)=TAoplus TB$ and $Tiota_X=iota_{TX}$ and $Tpi_X = pi_{TX}$ for the structure morphisms $iota_A,iota_B,pi_A,$ and $pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $Delta_A$ is the map such that $pi_{A,i}Delta_A=newcommandid{operatorname{id}}id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $TDelta_A$ satisfies $pi_{TA,i}T(Delta_A)=id_{TA}$ for $i=1,2$, thus $TDelta_A=Delta_{TA}$. A dual argument shows that $Tnabla_B=nabla_{TB}$, and finally $foplus g$ is defined to be the map such that
$$pi_{B,1}(foplus g)iota_{A,1} = f,$$
$$pi_{B,2}(foplus g)iota_{A,2} = g,$$
$$pi_{B,1}(foplus g)iota_{A,2} = 0,$$
and
$$pi_{B,2}(foplus g)iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(foplus g)$ satisfies the requirement to be $Tfoplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TAto TB to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 to 0 toby{id_0} 0 toby{id_0} 0.$$
This is clearly exact,
so
$$ 0 to T0 toby{id_{T0}} T0 toby{id_{T0}} T0$$
is exact, which implies that $$0=ker id_{T0} = T0=operatorname{im}_{id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $pi_Aiota_A =id_A$, so $(Tpi_A)(Tiota_A)=id_{TA}$. This also applies to $pi_Biota_B$, so $Tpi_B$ must be an epimorphism. Hence, exactness of
$$0to A toby{iota_A} Aoplus B toby{pi_B} B to 0$$
not only implies exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB, $$
but actually the exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB to 0,$$
and the fact that $(Tpi_A)(Tiota_A)=id_{TA}$ implies that this exact sequence is split, giving
$T(Aoplus B) simeq TAoplus TB$.
answered Jan 7 at 22:01
jgonjgon
13.7k22041
$endgroup$
add a comment |
$begingroup$
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : Ato B$ are morphisms, we define $f+g$ to be the composite
$$ Anewcommandtobyxrightarrowtoby{Delta_A} A oplus Atoby{foplus g}Boplus B toby{nabla_B} B,$$
where $Delta_A : Ato Aoplus A$ is the diagonal map and $nabla_B : Boplus Bto B$ is the codiagonal.
Then clearly any functor which satisfies $T(foplus g)=Tfoplus Tg$ and $TDelta_A = Delta_{TA}$ and $Tnabla_B = nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(Aoplus B)=TAoplus TB$ and $Tiota_X=iota_{TX}$ and $Tpi_X = pi_{TX}$ for the structure morphisms $iota_A,iota_B,pi_A,$ and $pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $Delta_A$ is the map such that $pi_{A,i}Delta_A=newcommandid{operatorname{id}}id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $TDelta_A$ satisfies $pi_{TA,i}T(Delta_A)=id_{TA}$ for $i=1,2$, thus $TDelta_A=Delta_{TA}$. A dual argument shows that $Tnabla_B=nabla_{TB}$, and finally $foplus g$ is defined to be the map such that
$$pi_{B,1}(foplus g)iota_{A,1} = f,$$
$$pi_{B,2}(foplus g)iota_{A,2} = g,$$
$$pi_{B,1}(foplus g)iota_{A,2} = 0,$$
and
$$pi_{B,2}(foplus g)iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(foplus g)$ satisfies the requirement to be $Tfoplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TAto TB to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 to 0 toby{id_0} 0 toby{id_0} 0.$$
This is clearly exact,
so
$$ 0 to T0 toby{id_{T0}} T0 toby{id_{T0}} T0$$
is exact, which implies that $$0=ker id_{T0} = T0=operatorname{im}_{id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $pi_Aiota_A =id_A$, so $(Tpi_A)(Tiota_A)=id_{TA}$. This also applies to $pi_Biota_B$, so $Tpi_B$ must be an epimorphism. Hence, exactness of
$$0to A toby{iota_A} Aoplus B toby{pi_B} B to 0$$
not only implies exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB, $$
but actually the exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB to 0,$$
and the fact that $(Tpi_A)(Tiota_A)=id_{TA}$ implies that this exact sequence is split, giving
$T(Aoplus B) simeq TAoplus TB$.
answered Jan 7 at 22:01
jgonjgon
13.7k22041
$endgroup$
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : Ato B$ are morphisms, we define $f+g$ to be the composite
$$ Anewcommandtobyxrightarrowtoby{Delta_A} A oplus Atoby{foplus g}Boplus B toby{nabla_B} B,$$
where $Delta_A : Ato Aoplus A$ is the diagonal map and $nabla_B : Boplus Bto B$ is the codiagonal.
Then clearly any functor which satisfies $T(foplus g)=Tfoplus Tg$ and $TDelta_A = Delta_{TA}$ and $Tnabla_B = nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(Aoplus B)=TAoplus TB$ and $Tiota_X=iota_{TX}$ and $Tpi_X = pi_{TX}$ for the structure morphisms $iota_A,iota_B,pi_A,$ and $pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $Delta_A$ is the map such that $pi_{A,i}Delta_A=newcommandid{operatorname{id}}id_A$ for $i=1,2$,
and then taking $T$ of this, we have that $TDelta_A$ satisfies $pi_{TA,i}T(Delta_A)=id_{TA}$ for $i=1,2$, thus $TDelta_A=Delta_{TA}$. A dual argument shows that $Tnabla_B=nabla_{TB}$, and finally $foplus g$ is defined to be the map such that
$$pi_{B,1}(foplus g)iota_{A,1} = f,$$
$$pi_{B,2}(foplus g)iota_{A,2} = g,$$
$$pi_{B,1}(foplus g)iota_{A,2} = 0,$$
and
$$pi_{B,2}(foplus g)iota_{A,1} = 0.$$
Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(foplus g)$ satisfies the requirement to be $Tfoplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TAto TB to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider
$$ 0 to 0 toby{id_0} 0 toby{id_0} 0.$$
This is clearly exact,
so
$$ 0 to T0 toby{id_{T0}} T0 toby{id_{T0}} T0$$
is exact, which implies that $$0=ker id_{T0} = T0=operatorname{im}_{id_{T0}},$$
as desired.
$T$ preserves the biproduct
Proof:
We know $pi_Aiota_A =id_A$, so $(Tpi_A)(Tiota_A)=id_{TA}$. This also applies to $pi_Biota_B$, so $Tpi_B$ must be an epimorphism. Hence, exactness of
$$0to A toby{iota_A} Aoplus B toby{pi_B} B to 0$$
not only implies exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB, $$
but actually the exactness of
$$0to TA toby{Tiota_A} T(Aoplus B) toby{Tpi_B} TB to 0,$$
and the fact that $(Tpi_A)(Tiota_A)=id_{TA}$ implies that this exact sequence is split, giving
$T(Aoplus B) simeq TAoplus TB$.
answered Jan 7 at 22:01
jgonjgon
13.7k22041
answered Jan 7 at 22:01
jgonjgon
13.7k22041
answered Jan 7 at 22:01
jgonjgon
13.7k22041
answered Jan 7 at 22:01
jgonjgon
13.7k22041
13.7k22041
add a comment |
add a comment |
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$begingroup$
A left exact functor is additive (me again for the same comment ;)). Rotman's definition is equivalent to any other. You proved that a left exact functor preserve finite direct sums. If a functor preserve the zero object and direct sums, then it is additive. This is because the addition of morphisms $f,g:Ato B$ is $Aoverset{Delta}longrightarrow Aoplus Aoverset{foplus g}longrightarrow Boplus Boverset{nabla}longrightarrow B$ where $Delta$ is the diagonal and $nabla$ the codiagonal.
$endgroup$
– Roland
Jan 7 at 20:54
$begingroup$
@Roland Good point, I forgot about that. Let me see if I can work out the details there.
$endgroup$
– jgon
Jan 7 at 20:58
$begingroup$
@Roland I've added an answer working out the details, but if you post your comment as an answer, then I'll accept that, since it did completely answer my question :)
$endgroup$
– jgon
Jan 7 at 22:02
$begingroup$
That's ok, you wrote the details, I don't see the point of adding another answer.
$endgroup$
– Roland
Jan 7 at 22:10