Mixing
About a specific step in a proof of the fact that filtered colimits and finite limits commute in...
$begingroup$
I'm currently working on the following theorem from Emily Riehl's Category Theory in Context:
Theorem 3.8.9. Filtered colimits commute with finite limits in $mathbf{Set}$.
I understand most of the proof, except for a small detail which I will proceed to explain next. The proof starts by considering the canonical map
$$
kappa : text{colim}_J text{lim}_IF(i,j) longrightarrow text{lim}_Itext{colim}_JF(i,j) tag{1}
$$
and trying to prove that it is a bijection. The author also previosuly highlights a key remark, which can be paraphrased as:
When $J$ is a small filtered category, for any functor $G in mathbf{Set}^J$ we have that $text{colim}_JG = big(coprod_j Gjbig) / sim$ with $x in Gj sim y in Gk$ if and only if we have arrows $f : j to t, g : k to t$ such that $Gf(x) = Gg(y)$.
To see that $kappa$ is surjective, we consider an element of $text{lim}_Itext{colim}_JF(i,j)$. Since we are working with the limit of a set valued funtor, this is equivalent to choosing a cone $(lambda : 1 Rightarrow text{colim}_JF(i,-))_{iin I}$. Each $lambda_i$ corresponds by the remark to an element of $text{colim}_JF(i,-)$ which can be identified as the class of equivalence of a certain element $lambda_i in F(i,j_i)$. Now, the author claims that we can find $t in J$ so that each $lambda_i$ is equivalent to some $lambda'_i in F(i,t)$ and that moreover, $(lambda': 1 Rightarrow F(-,t))_{i in I}$ is a cone.
The first part of this claim, I do have managed to prove: since $I$ is finite, the full subcategory $J_I$ spanned by the $(j_i)_i$ can be thought as a diagram $I to J$. The latter is filtered and $I$ is finite, so it follows that we have a cone $(mu_i: j_i to t)_{i in I}$ under the diagram. Defining $lambda'_i$ to be $F(1_i,mu_i)(lambda_i)$, we see that $lambda_i sim lambda'_i$ via the arrows $1_{t}$ and $mu_i$.
From here on, I have unable to finish the proof of the claim, that is, I haven't been able to prove that $lambda'$ is a cone. I have tried to write maps $lambda'_i : 1 to F(i,t) $ as compositions $F(1_i,mu_i)lambda_i$ and then using that $lambda$ is itself a cone (on colimit objects), but I have not been able to deal with the fact that these factorizations change the object in $J$ (for example, to $j_i$ and $j_{i'}$, if we deal with an arrow $g : ito i'$ to begin with).
Any ideas on how to conclude from here? Thanks in advance.
category-theory proof-explanation limits-colimits
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
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add a comment |
$begingroup$
I'm currently working on the following theorem from Emily Riehl's Category Theory in Context:
Theorem 3.8.9. Filtered colimits commute with finite limits in $mathbf{Set}$.
I understand most of the proof, except for a small detail which I will proceed to explain next. The proof starts by considering the canonical map
$$
kappa : text{colim}_J text{lim}_IF(i,j) longrightarrow text{lim}_Itext{colim}_JF(i,j) tag{1}
$$
and trying to prove that it is a bijection. The author also previosuly highlights a key remark, which can be paraphrased as:
When $J$ is a small filtered category, for any functor $G in mathbf{Set}^J$ we have that $text{colim}_JG = big(coprod_j Gjbig) / sim$ with $x in Gj sim y in Gk$ if and only if we have arrows $f : j to t, g : k to t$ such that $Gf(x) = Gg(y)$.
To see that $kappa$ is surjective, we consider an element of $text{lim}_Itext{colim}_JF(i,j)$. Since we are working with the limit of a set valued funtor, this is equivalent to choosing a cone $(lambda : 1 Rightarrow text{colim}_JF(i,-))_{iin I}$. Each $lambda_i$ corresponds by the remark to an element of $text{colim}_JF(i,-)$ which can be identified as the class of equivalence of a certain element $lambda_i in F(i,j_i)$. Now, the author claims that we can find $t in J$ so that each $lambda_i$ is equivalent to some $lambda'_i in F(i,t)$ and that moreover, $(lambda': 1 Rightarrow F(-,t))_{i in I}$ is a cone.
The first part of this claim, I do have managed to prove: since $I$ is finite, the full subcategory $J_I$ spanned by the $(j_i)_i$ can be thought as a diagram $I to J$. The latter is filtered and $I$ is finite, so it follows that we have a cone $(mu_i: j_i to t)_{i in I}$ under the diagram. Defining $lambda'_i$ to be $F(1_i,mu_i)(lambda_i)$, we see that $lambda_i sim lambda'_i$ via the arrows $1_{t}$ and $mu_i$.
From here on, I have unable to finish the proof of the claim, that is, I haven't been able to prove that $lambda'$ is a cone. I have tried to write maps $lambda'_i : 1 to F(i,t) $ as compositions $F(1_i,mu_i)lambda_i$ and then using that $lambda$ is itself a cone (on colimit objects), but I have not been able to deal with the fact that these factorizations change the object in $J$ (for example, to $j_i$ and $j_{i'}$, if we deal with an arrow $g : ito i'$ to begin with).
Any ideas on how to conclude from here? Thanks in advance.
category-theory proof-explanation limits-colimits
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
$endgroup$
add a comment |
$begingroup$
I'm currently working on the following theorem from Emily Riehl's Category Theory in Context:
Theorem 3.8.9. Filtered colimits commute with finite limits in $mathbf{Set}$.
I understand most of the proof, except for a small detail which I will proceed to explain next. The proof starts by considering the canonical map
$$
kappa : text{colim}_J text{lim}_IF(i,j) longrightarrow text{lim}_Itext{colim}_JF(i,j) tag{1}
$$
and trying to prove that it is a bijection. The author also previosuly highlights a key remark, which can be paraphrased as:
When $J$ is a small filtered category, for any functor $G in mathbf{Set}^J$ we have that $text{colim}_JG = big(coprod_j Gjbig) / sim$ with $x in Gj sim y in Gk$ if and only if we have arrows $f : j to t, g : k to t$ such that $Gf(x) = Gg(y)$.
To see that $kappa$ is surjective, we consider an element of $text{lim}_Itext{colim}_JF(i,j)$. Since we are working with the limit of a set valued funtor, this is equivalent to choosing a cone $(lambda : 1 Rightarrow text{colim}_JF(i,-))_{iin I}$. Each $lambda_i$ corresponds by the remark to an element of $text{colim}_JF(i,-)$ which can be identified as the class of equivalence of a certain element $lambda_i in F(i,j_i)$. Now, the author claims that we can find $t in J$ so that each $lambda_i$ is equivalent to some $lambda'_i in F(i,t)$ and that moreover, $(lambda': 1 Rightarrow F(-,t))_{i in I}$ is a cone.
The first part of this claim, I do have managed to prove: since $I$ is finite, the full subcategory $J_I$ spanned by the $(j_i)_i$ can be thought as a diagram $I to J$. The latter is filtered and $I$ is finite, so it follows that we have a cone $(mu_i: j_i to t)_{i in I}$ under the diagram. Defining $lambda'_i$ to be $F(1_i,mu_i)(lambda_i)$, we see that $lambda_i sim lambda'_i$ via the arrows $1_{t}$ and $mu_i$.
From here on, I have unable to finish the proof of the claim, that is, I haven't been able to prove that $lambda'$ is a cone. I have tried to write maps $lambda'_i : 1 to F(i,t) $ as compositions $F(1_i,mu_i)lambda_i$ and then using that $lambda$ is itself a cone (on colimit objects), but I have not been able to deal with the fact that these factorizations change the object in $J$ (for example, to $j_i$ and $j_{i'}$, if we deal with an arrow $g : ito i'$ to begin with).
Any ideas on how to conclude from here? Thanks in advance.
category-theory proof-explanation limits-colimits
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
$endgroup$
I'm currently working on the following theorem from Emily Riehl's Category Theory in Context:
Theorem 3.8.9. Filtered colimits commute with finite limits in $mathbf{Set}$.
I understand most of the proof, except for a small detail which I will proceed to explain next. The proof starts by considering the canonical map
$$
kappa : text{colim}_J text{lim}_IF(i,j) longrightarrow text{lim}_Itext{colim}_JF(i,j) tag{1}
$$
and trying to prove that it is a bijection. The author also previosuly highlights a key remark, which can be paraphrased as:
When $J$ is a small filtered category, for any functor $G in mathbf{Set}^J$ we have that $text{colim}_JG = big(coprod_j Gjbig) / sim$ with $x in Gj sim y in Gk$ if and only if we have arrows $f : j to t, g : k to t$ such that $Gf(x) = Gg(y)$.
To see that $kappa$ is surjective, we consider an element of $text{lim}_Itext{colim}_JF(i,j)$. Since we are working with the limit of a set valued funtor, this is equivalent to choosing a cone $(lambda : 1 Rightarrow text{colim}_JF(i,-))_{iin I}$. Each $lambda_i$ corresponds by the remark to an element of $text{colim}_JF(i,-)$ which can be identified as the class of equivalence of a certain element $lambda_i in F(i,j_i)$. Now, the author claims that we can find $t in J$ so that each $lambda_i$ is equivalent to some $lambda'_i in F(i,t)$ and that moreover, $(lambda': 1 Rightarrow F(-,t))_{i in I}$ is a cone.
The first part of this claim, I do have managed to prove: since $I$ is finite, the full subcategory $J_I$ spanned by the $(j_i)_i$ can be thought as a diagram $I to J$. The latter is filtered and $I$ is finite, so it follows that we have a cone $(mu_i: j_i to t)_{i in I}$ under the diagram. Defining $lambda'_i$ to be $F(1_i,mu_i)(lambda_i)$, we see that $lambda_i sim lambda'_i$ via the arrows $1_{t}$ and $mu_i$.
From here on, I have unable to finish the proof of the claim, that is, I haven't been able to prove that $lambda'$ is a cone. I have tried to write maps $lambda'_i : 1 to F(i,t) $ as compositions $F(1_i,mu_i)lambda_i$ and then using that $lambda$ is itself a cone (on colimit objects), but I have not been able to deal with the fact that these factorizations change the object in $J$ (for example, to $j_i$ and $j_{i'}$, if we deal with an arrow $g : ito i'$ to begin with).
Any ideas on how to conclude from here? Thanks in advance.
category-theory proof-explanation limits-colimits
category-theory proof-explanation limits-colimits
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
edited Jan 16 at 10:36
Guido A.
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
asked Jan 16 at 9:36
Guido A.Guido A.
7,5161730
7,5161730
add a comment |
add a comment |
1 Answer
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$begingroup$
For any functor of the form $U:mathcal Atimesmathcal Btomathcal C$ and any morhpisms $alpha:ato a', beta:bto b'$, we have the following symmetry:
$$U(alpha,b'),U(a,beta) = U(alpha,beta) = U(a',beta),U(alpha,b) ,$$
where e.g. $U(alpha,b)$ stands for $U(alpha,1_b)$
simply because we already have $(alpha,1_{b'})(1_a,beta)=(alpha,beta)=(1_{a'},beta)(alpha,1_b)$ in $mathcal Atimesmathcal B$.
Now, let $gamma:ito i'$ be an arrow in $I$, then we have
$$F(gamma,t),lambda_i' = F(gamma,t),F(i,mu_i),lambda_i = F(gamma,mu_i),lambda_i =\
= F(i',mu_i),F(gamma,j_i),lambda_i overset{lambdatext{ cone}}= F(i',mu_i),lambda_{i'} = lambda'_{i'} $$
answered Jan 18 at 1:30
BerciBerci
61k23674
$endgroup$
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
For any functor of the form $U:mathcal Atimesmathcal Btomathcal C$ and any morhpisms $alpha:ato a', beta:bto b'$, we have the following symmetry:
$$U(alpha,b'),U(a,beta) = U(alpha,beta) = U(a',beta),U(alpha,b) ,$$
where e.g. $U(alpha,b)$ stands for $U(alpha,1_b)$
simply because we already have $(alpha,1_{b'})(1_a,beta)=(alpha,beta)=(1_{a'},beta)(alpha,1_b)$ in $mathcal Atimesmathcal B$.
Now, let $gamma:ito i'$ be an arrow in $I$, then we have
$$F(gamma,t),lambda_i' = F(gamma,t),F(i,mu_i),lambda_i = F(gamma,mu_i),lambda_i =\
= F(i',mu_i),F(gamma,j_i),lambda_i overset{lambdatext{ cone}}= F(i',mu_i),lambda_{i'} = lambda'_{i'} $$
answered Jan 18 at 1:30
BerciBerci
61k23674
$endgroup$
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
add a comment |
$begingroup$
For any functor of the form $U:mathcal Atimesmathcal Btomathcal C$ and any morhpisms $alpha:ato a', beta:bto b'$, we have the following symmetry:
$$U(alpha,b'),U(a,beta) = U(alpha,beta) = U(a',beta),U(alpha,b) ,$$
where e.g. $U(alpha,b)$ stands for $U(alpha,1_b)$
simply because we already have $(alpha,1_{b'})(1_a,beta)=(alpha,beta)=(1_{a'},beta)(alpha,1_b)$ in $mathcal Atimesmathcal B$.
Now, let $gamma:ito i'$ be an arrow in $I$, then we have
$$F(gamma,t),lambda_i' = F(gamma,t),F(i,mu_i),lambda_i = F(gamma,mu_i),lambda_i =\
= F(i',mu_i),F(gamma,j_i),lambda_i overset{lambdatext{ cone}}= F(i',mu_i),lambda_{i'} = lambda'_{i'} $$
answered Jan 18 at 1:30
BerciBerci
61k23674
$endgroup$
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
add a comment |
$begingroup$
For any functor of the form $U:mathcal Atimesmathcal Btomathcal C$ and any morhpisms $alpha:ato a', beta:bto b'$, we have the following symmetry:
$$U(alpha,b'),U(a,beta) = U(alpha,beta) = U(a',beta),U(alpha,b) ,$$
where e.g. $U(alpha,b)$ stands for $U(alpha,1_b)$
simply because we already have $(alpha,1_{b'})(1_a,beta)=(alpha,beta)=(1_{a'},beta)(alpha,1_b)$ in $mathcal Atimesmathcal B$.
Now, let $gamma:ito i'$ be an arrow in $I$, then we have
$$F(gamma,t),lambda_i' = F(gamma,t),F(i,mu_i),lambda_i = F(gamma,mu_i),lambda_i =\
= F(i',mu_i),F(gamma,j_i),lambda_i overset{lambdatext{ cone}}= F(i',mu_i),lambda_{i'} = lambda'_{i'} $$
answered Jan 18 at 1:30
BerciBerci
61k23674
$endgroup$
For any functor of the form $U:mathcal Atimesmathcal Btomathcal C$ and any morhpisms $alpha:ato a', beta:bto b'$, we have the following symmetry:
$$U(alpha,b'),U(a,beta) = U(alpha,beta) = U(a',beta),U(alpha,b) ,$$
where e.g. $U(alpha,b)$ stands for $U(alpha,1_b)$
simply because we already have $(alpha,1_{b'})(1_a,beta)=(alpha,beta)=(1_{a'},beta)(alpha,1_b)$ in $mathcal Atimesmathcal B$.
Now, let $gamma:ito i'$ be an arrow in $I$, then we have
$$F(gamma,t),lambda_i' = F(gamma,t),F(i,mu_i),lambda_i = F(gamma,mu_i),lambda_i =\
= F(i',mu_i),F(gamma,j_i),lambda_i overset{lambdatext{ cone}}= F(i',mu_i),lambda_{i'} = lambda'_{i'} $$
answered Jan 18 at 1:30
BerciBerci
61k23674
answered Jan 18 at 1:30
BerciBerci
61k23674
answered Jan 18 at 1:30
BerciBerci
61k23674
answered Jan 18 at 1:30
BerciBerci
61k23674
61k23674
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
add a comment |
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
$begingroup$
Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it.
$endgroup$
– Guido A.
Jan 18 at 17:34
1
1
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
$begingroup$
The part with the $Ito J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it..
$endgroup$
– Berci
Jan 18 at 18:30
add a comment |
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