Can it be useful to think of functors as representing themselves ?
$begingroup$
Here's a thought I had and I wonder if it can be of any use, for instance has it ever helped proving any result (however minor the result).
Say you're in a situation where you have some objects in a category $C$, and some universal problem (i.e. functor $F: C^{op} to mathbf{Set}$) comes up, and you'd like it to have a solution (i.e. a representative of said functor), because if it did, then you could perform such and such arguments which would then allow you to prove something concerning your original situation (not involving the solution to the universal problem !). However, you're in bad luck, and it so happens that your universal problem has no solution: the functor is not representable.
No big deal, your arguments were abstract enough that you can look at the full subcategory $C[F]$ of $widehat{C}$ (presheaves on $C$) on representable presheaves and $F$. In this category, by the Yoneda lemma, the restriction of $hom(-,F)$ to the representable presheaves is isomorphic to $F$ (by transfer along the Yoneda embedding), so we have an extension of $F$ to $C[F]$ that is isomorphic to $F$ on the subcategory where it makes sense. In some sense, we've added a representative of $F$ to $C$ (hence my ad hoc notation). Then we can perform the earlier arguments in $C[F]$ because they only involved objects of $C$ and the representative of $F$ (which is $F$ here), and then we get the result on our objects, said result not involving the solution to the universal problem, we can (if the result is nice enough) pull it back to $C$ along the Yoneda embedding.
Thus, even though our argument relied on finding a representative to a functor, we could perform it, simply by enlarging our category for a moment.
In my mind, this is the same kind of spirit as proving properties of reals by adding $i$, getting to $mathbb{C}$ and then coming back to $mathbb{R}$, or more subtle yet, proving results concerning a model of ZFC $V$ by forcing with a generic $G$ and then forgetting about it (a bit less subtle : this happens often in model theory where you enlarge the language and then come back to the original language) : so the spirit behind this idea is well-known, but I wonder if something along the lines of the situation I mention ever happens, if it can actually come in handy. My question is a bit vague so I hope it's suitable.
soft-question category-theory universal-property representable-functor
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
$endgroup$
add a comment |
$begingroup$
Here's a thought I had and I wonder if it can be of any use, for instance has it ever helped proving any result (however minor the result).
Say you're in a situation where you have some objects in a category $C$, and some universal problem (i.e. functor $F: C^{op} to mathbf{Set}$) comes up, and you'd like it to have a solution (i.e. a representative of said functor), because if it did, then you could perform such and such arguments which would then allow you to prove something concerning your original situation (not involving the solution to the universal problem !). However, you're in bad luck, and it so happens that your universal problem has no solution: the functor is not representable.
No big deal, your arguments were abstract enough that you can look at the full subcategory $C[F]$ of $widehat{C}$ (presheaves on $C$) on representable presheaves and $F$. In this category, by the Yoneda lemma, the restriction of $hom(-,F)$ to the representable presheaves is isomorphic to $F$ (by transfer along the Yoneda embedding), so we have an extension of $F$ to $C[F]$ that is isomorphic to $F$ on the subcategory where it makes sense. In some sense, we've added a representative of $F$ to $C$ (hence my ad hoc notation). Then we can perform the earlier arguments in $C[F]$ because they only involved objects of $C$ and the representative of $F$ (which is $F$ here), and then we get the result on our objects, said result not involving the solution to the universal problem, we can (if the result is nice enough) pull it back to $C$ along the Yoneda embedding.
Thus, even though our argument relied on finding a representative to a functor, we could perform it, simply by enlarging our category for a moment.
In my mind, this is the same kind of spirit as proving properties of reals by adding $i$, getting to $mathbb{C}$ and then coming back to $mathbb{R}$, or more subtle yet, proving results concerning a model of ZFC $V$ by forcing with a generic $G$ and then forgetting about it (a bit less subtle : this happens often in model theory where you enlarge the language and then come back to the original language) : so the spirit behind this idea is well-known, but I wonder if something along the lines of the situation I mention ever happens, if it can actually come in handy. My question is a bit vague so I hope it's suitable.
soft-question category-theory universal-property representable-functor
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
$endgroup$
$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
1
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07
add a comment |
$begingroup$
Here's a thought I had and I wonder if it can be of any use, for instance has it ever helped proving any result (however minor the result).
Say you're in a situation where you have some objects in a category $C$, and some universal problem (i.e. functor $F: C^{op} to mathbf{Set}$) comes up, and you'd like it to have a solution (i.e. a representative of said functor), because if it did, then you could perform such and such arguments which would then allow you to prove something concerning your original situation (not involving the solution to the universal problem !). However, you're in bad luck, and it so happens that your universal problem has no solution: the functor is not representable.
No big deal, your arguments were abstract enough that you can look at the full subcategory $C[F]$ of $widehat{C}$ (presheaves on $C$) on representable presheaves and $F$. In this category, by the Yoneda lemma, the restriction of $hom(-,F)$ to the representable presheaves is isomorphic to $F$ (by transfer along the Yoneda embedding), so we have an extension of $F$ to $C[F]$ that is isomorphic to $F$ on the subcategory where it makes sense. In some sense, we've added a representative of $F$ to $C$ (hence my ad hoc notation). Then we can perform the earlier arguments in $C[F]$ because they only involved objects of $C$ and the representative of $F$ (which is $F$ here), and then we get the result on our objects, said result not involving the solution to the universal problem, we can (if the result is nice enough) pull it back to $C$ along the Yoneda embedding.
Thus, even though our argument relied on finding a representative to a functor, we could perform it, simply by enlarging our category for a moment.
In my mind, this is the same kind of spirit as proving properties of reals by adding $i$, getting to $mathbb{C}$ and then coming back to $mathbb{R}$, or more subtle yet, proving results concerning a model of ZFC $V$ by forcing with a generic $G$ and then forgetting about it (a bit less subtle : this happens often in model theory where you enlarge the language and then come back to the original language) : so the spirit behind this idea is well-known, but I wonder if something along the lines of the situation I mention ever happens, if it can actually come in handy. My question is a bit vague so I hope it's suitable.
soft-question category-theory universal-property representable-functor
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
$endgroup$
Here's a thought I had and I wonder if it can be of any use, for instance has it ever helped proving any result (however minor the result).
Say you're in a situation where you have some objects in a category $C$, and some universal problem (i.e. functor $F: C^{op} to mathbf{Set}$) comes up, and you'd like it to have a solution (i.e. a representative of said functor), because if it did, then you could perform such and such arguments which would then allow you to prove something concerning your original situation (not involving the solution to the universal problem !). However, you're in bad luck, and it so happens that your universal problem has no solution: the functor is not representable.
No big deal, your arguments were abstract enough that you can look at the full subcategory $C[F]$ of $widehat{C}$ (presheaves on $C$) on representable presheaves and $F$. In this category, by the Yoneda lemma, the restriction of $hom(-,F)$ to the representable presheaves is isomorphic to $F$ (by transfer along the Yoneda embedding), so we have an extension of $F$ to $C[F]$ that is isomorphic to $F$ on the subcategory where it makes sense. In some sense, we've added a representative of $F$ to $C$ (hence my ad hoc notation). Then we can perform the earlier arguments in $C[F]$ because they only involved objects of $C$ and the representative of $F$ (which is $F$ here), and then we get the result on our objects, said result not involving the solution to the universal problem, we can (if the result is nice enough) pull it back to $C$ along the Yoneda embedding.
Thus, even though our argument relied on finding a representative to a functor, we could perform it, simply by enlarging our category for a moment.
In my mind, this is the same kind of spirit as proving properties of reals by adding $i$, getting to $mathbb{C}$ and then coming back to $mathbb{R}$, or more subtle yet, proving results concerning a model of ZFC $V$ by forcing with a generic $G$ and then forgetting about it (a bit less subtle : this happens often in model theory where you enlarge the language and then come back to the original language) : so the spirit behind this idea is well-known, but I wonder if something along the lines of the situation I mention ever happens, if it can actually come in handy. My question is a bit vague so I hope it's suitable.
soft-question category-theory universal-property representable-functor
soft-question category-theory universal-property representable-functor
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
asked Dec 4 '18 at 14:15
MaxMax
13.8k11142
13.8k11142
$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
1
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07
add a comment |
$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
1
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07
$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
1
1
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Here's a simple example: suppose you want to compute the monoid of endomorphisms of $F$. If $F$ is representable by an object $c$, by the Yoneda lemma you could do this computation by computing the monoid of endomorphisms of $c$. This is frequently useful. If $F$ isn't representable, then the construction you suggest tells you nothing: just that the monoid of endomorphisms of $F$ is the monoid of endomorphisms of $F$, which you already knew.
On the other hand, $F$ can sometimes be "almost representable" in a useful way, by a pro-object or an ind-object depending on whether $F$ is covariant or contravariant. See this blog post for details.
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
add a comment |
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$begingroup$
Here's a simple example: suppose you want to compute the monoid of endomorphisms of $F$. If $F$ is representable by an object $c$, by the Yoneda lemma you could do this computation by computing the monoid of endomorphisms of $c$. This is frequently useful. If $F$ isn't representable, then the construction you suggest tells you nothing: just that the monoid of endomorphisms of $F$ is the monoid of endomorphisms of $F$, which you already knew.
On the other hand, $F$ can sometimes be "almost representable" in a useful way, by a pro-object or an ind-object depending on whether $F$ is covariant or contravariant. See this blog post for details.
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
add a comment |
$begingroup$
Here's a simple example: suppose you want to compute the monoid of endomorphisms of $F$. If $F$ is representable by an object $c$, by the Yoneda lemma you could do this computation by computing the monoid of endomorphisms of $c$. This is frequently useful. If $F$ isn't representable, then the construction you suggest tells you nothing: just that the monoid of endomorphisms of $F$ is the monoid of endomorphisms of $F$, which you already knew.
On the other hand, $F$ can sometimes be "almost representable" in a useful way, by a pro-object or an ind-object depending on whether $F$ is covariant or contravariant. See this blog post for details.
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
add a comment |
$begingroup$
Here's a simple example: suppose you want to compute the monoid of endomorphisms of $F$. If $F$ is representable by an object $c$, by the Yoneda lemma you could do this computation by computing the monoid of endomorphisms of $c$. This is frequently useful. If $F$ isn't representable, then the construction you suggest tells you nothing: just that the monoid of endomorphisms of $F$ is the monoid of endomorphisms of $F$, which you already knew.
On the other hand, $F$ can sometimes be "almost representable" in a useful way, by a pro-object or an ind-object depending on whether $F$ is covariant or contravariant. See this blog post for details.
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
$endgroup$
Here's a simple example: suppose you want to compute the monoid of endomorphisms of $F$. If $F$ is representable by an object $c$, by the Yoneda lemma you could do this computation by computing the monoid of endomorphisms of $c$. This is frequently useful. If $F$ isn't representable, then the construction you suggest tells you nothing: just that the monoid of endomorphisms of $F$ is the monoid of endomorphisms of $F$, which you already knew.
On the other hand, $F$ can sometimes be "almost representable" in a useful way, by a pro-object or an ind-object depending on whether $F$ is covariant or contravariant. See this blog post for details.
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
edited Dec 5 '18 at 1:32
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
answered Dec 4 '18 at 20:21
Qiaochu YuanQiaochu Yuan
278k32585921
278k32585921
add a comment |
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$begingroup$
Excuse me if I am misunderstanding, but $F$ is still not representable in $C[F]$ is it?
$endgroup$
– asdq
Dec 4 '18 at 14:56
$begingroup$
@asdq : $F$ is not defined on the whole of $C[F]$ so it doesn't make sense to ask if it's representable. But $hom(-,F)$ is a functor on $C[F]$ that is (clearly) representable and its restriction to $C$ is isomorphic to $F$
$endgroup$
– Max
Dec 4 '18 at 15:50
$begingroup$
I see, but this remains true of you just take the entire presheaf category. Of course, a lot of times this can be really useful. An example that comes to my mind is the theory of ind-objects, where one uses the cocompleteness of the presheaf category in order to obtain inductive limits that fail to exist in the original category.
$endgroup$
– asdq
Dec 4 '18 at 16:36
$begingroup$
@asdq : sure, but if $C$ is large, the whole presheaf category might be too large and I was trying to avoid this kind of problem by taking $C[F]$. Thank you for mentioning ind-objects, I'll take a look at it. If you have the time and feel like it, perhaps you could write some more details about it in an answer ?
$endgroup$
– Max
Dec 4 '18 at 17:32
1
$begingroup$
@Max It may make sense to focus on the free cocompletion aspect of presheaves when considering "presheaves" on large categories. That is, the category of small presheaves as defined in Limits of small functors.
$endgroup$
– Derek Elkins
Dec 4 '18 at 18:07