Mixing
Equivalence of Categories does not preserve enrichment
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I am learning some category theory currently, and was thinking about the following problem: Suppose $F: C to D$ is a functor that induces an equivalence of categories and that $C,D$ are enriched over some category $M$. Then does $F$ preserve the enriched structure, that is is $Hom(c,c') simeq Hom(F(c),F(c'))$, where the isomorphism is in the category $M$. We only know that this isomorphism is an isomorphism of sets, and my intuition tells me that we should not get a natural isomorphism in $M$. However, I tried finding some counter examples with categories enriched over familiar things like topological spaces, and abelian groups, but could not find a counter example. Any ideas?
abstract-algebra category-theory
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
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add a comment |
$begingroup$
I am learning some category theory currently, and was thinking about the following problem: Suppose $F: C to D$ is a functor that induces an equivalence of categories and that $C,D$ are enriched over some category $M$. Then does $F$ preserve the enriched structure, that is is $Hom(c,c') simeq Hom(F(c),F(c'))$, where the isomorphism is in the category $M$. We only know that this isomorphism is an isomorphism of sets, and my intuition tells me that we should not get a natural isomorphism in $M$. However, I tried finding some counter examples with categories enriched over familiar things like topological spaces, and abelian groups, but could not find a counter example. Any ideas?
abstract-algebra category-theory
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
$endgroup$
add a comment |
$begingroup$
I am learning some category theory currently, and was thinking about the following problem: Suppose $F: C to D$ is a functor that induces an equivalence of categories and that $C,D$ are enriched over some category $M$. Then does $F$ preserve the enriched structure, that is is $Hom(c,c') simeq Hom(F(c),F(c'))$, where the isomorphism is in the category $M$. We only know that this isomorphism is an isomorphism of sets, and my intuition tells me that we should not get a natural isomorphism in $M$. However, I tried finding some counter examples with categories enriched over familiar things like topological spaces, and abelian groups, but could not find a counter example. Any ideas?
abstract-algebra category-theory
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
$endgroup$
I am learning some category theory currently, and was thinking about the following problem: Suppose $F: C to D$ is a functor that induces an equivalence of categories and that $C,D$ are enriched over some category $M$. Then does $F$ preserve the enriched structure, that is is $Hom(c,c') simeq Hom(F(c),F(c'))$, where the isomorphism is in the category $M$. We only know that this isomorphism is an isomorphism of sets, and my intuition tells me that we should not get a natural isomorphism in $M$. However, I tried finding some counter examples with categories enriched over familiar things like topological spaces, and abelian groups, but could not find a counter example. Any ideas?
abstract-algebra category-theory
abstract-algebra category-theory
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
asked Feb 1 at 18:48
Sheel StueberSheel Stueber
691210
691210
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, certainly not. For instance, let $R$ and $S$ be rings and $f:Rto S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=mathbb{Z}$ and $f:mathbb{Z}tomathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).
An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:Ato B$ is the composition of $(f,g):Ato Boplus B$ with the fold map $Boplus Bto B$.
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
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add a comment |
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Eric Wofsey has already given a great answer (+1) giving an example of two equivalent categories enriched over abelian groups where the abelian groups fail to be preserved and also added a good point about the preservation of enrichment in additive categories.
However, I'd like to add an additional point.
Enriched categories are not (necessarily) ordinary categories. In fact, I'd like to just say that they are not ordinary categories. If the enriching category, $V$, is concrete, as is the case with $mathbf{Top}$ or $mathbf{Ab}$ enriched categories then your enriched category can be thought of as having a natural ordinary category structure as well. However, identifying the ordinary category with the enriched category would be a slight abuse of notation, since the hom objects in the enriched category are objects of $V$ and the hom objects of the associated ordinary category are objects of $mathbf{Set}$. That's generally ok, since we usually speak of abelian groups and topological spaces as being sets.
However, this distinction between $V$-enriched categories and their associated ordinary categories (when $V$ is concrete) is very relevant here, because one should be careful about what we mean by an equivalence of categories.
If $C$ and $D$ are $V$-enriched categories, then an equivalence of those categories should be a pair of $V$-functors $F:Cto D$ and $G:Dto C$ with $Fcirc G simeq 1_D$ and $Gcirc F simeq 1_C$ (the natural isomorphisms here should also be $V$-natural isomorphisms). With this notion of equivalence, an equivalence between $C$ and $D$ would naturally preserve the enrichment, and this is the correct notion of an equivalence between $C$ and $D$.
As Eric Wofsey has shown with his example, not only is there no reason to think that an equivalence between the associated ordinary categories would induce an equivalence between the original $V$-enriched categories, it is not true. The main point of my answer is that there aren't even necessarily ordinary categories associated to enriched categories, since your hom objects don't necessarily even lie in a concrete category (Yes I know we can always take the generalized elements of the unit to get an ordinary category from an enriched category, but that's beside the point here).
answered Feb 1 at 23:56
jgonjgon
16.5k32143
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
No, certainly not. For instance, let $R$ and $S$ be rings and $f:Rto S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=mathbb{Z}$ and $f:mathbb{Z}tomathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).
An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:Ato B$ is the composition of $(f,g):Ato Boplus B$ with the fold map $Boplus Bto B$.
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
$endgroup$
add a comment |
$begingroup$
No, certainly not. For instance, let $R$ and $S$ be rings and $f:Rto S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=mathbb{Z}$ and $f:mathbb{Z}tomathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).
An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:Ato B$ is the composition of $(f,g):Ato Boplus B$ with the fold map $Boplus Bto B$.
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
$endgroup$
add a comment |
$begingroup$
No, certainly not. For instance, let $R$ and $S$ be rings and $f:Rto S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=mathbb{Z}$ and $f:mathbb{Z}tomathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).
An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:Ato B$ is the composition of $(f,g):Ato Boplus B$ with the fold map $Boplus Bto B$.
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
$endgroup$
No, certainly not. For instance, let $R$ and $S$ be rings and $f:Rto S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=mathbb{Z}$ and $f:mathbb{Z}tomathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).
An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:Ato B$ is the composition of $(f,g):Ato Boplus B$ with the fold map $Boplus Bto B$.
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 22:10
Eric WofseyEric Wofsey
193k14220352
193k14220352
add a comment |
add a comment |
$begingroup$
Eric Wofsey has already given a great answer (+1) giving an example of two equivalent categories enriched over abelian groups where the abelian groups fail to be preserved and also added a good point about the preservation of enrichment in additive categories.
However, I'd like to add an additional point.
Enriched categories are not (necessarily) ordinary categories. In fact, I'd like to just say that they are not ordinary categories. If the enriching category, $V$, is concrete, as is the case with $mathbf{Top}$ or $mathbf{Ab}$ enriched categories then your enriched category can be thought of as having a natural ordinary category structure as well. However, identifying the ordinary category with the enriched category would be a slight abuse of notation, since the hom objects in the enriched category are objects of $V$ and the hom objects of the associated ordinary category are objects of $mathbf{Set}$. That's generally ok, since we usually speak of abelian groups and topological spaces as being sets.
However, this distinction between $V$-enriched categories and their associated ordinary categories (when $V$ is concrete) is very relevant here, because one should be careful about what we mean by an equivalence of categories.
If $C$ and $D$ are $V$-enriched categories, then an equivalence of those categories should be a pair of $V$-functors $F:Cto D$ and $G:Dto C$ with $Fcirc G simeq 1_D$ and $Gcirc F simeq 1_C$ (the natural isomorphisms here should also be $V$-natural isomorphisms). With this notion of equivalence, an equivalence between $C$ and $D$ would naturally preserve the enrichment, and this is the correct notion of an equivalence between $C$ and $D$.
As Eric Wofsey has shown with his example, not only is there no reason to think that an equivalence between the associated ordinary categories would induce an equivalence between the original $V$-enriched categories, it is not true. The main point of my answer is that there aren't even necessarily ordinary categories associated to enriched categories, since your hom objects don't necessarily even lie in a concrete category (Yes I know we can always take the generalized elements of the unit to get an ordinary category from an enriched category, but that's beside the point here).
answered Feb 1 at 23:56
jgonjgon
16.5k32143
$endgroup$
add a comment |
$begingroup$
Eric Wofsey has already given a great answer (+1) giving an example of two equivalent categories enriched over abelian groups where the abelian groups fail to be preserved and also added a good point about the preservation of enrichment in additive categories.
However, I'd like to add an additional point.
Enriched categories are not (necessarily) ordinary categories. In fact, I'd like to just say that they are not ordinary categories. If the enriching category, $V$, is concrete, as is the case with $mathbf{Top}$ or $mathbf{Ab}$ enriched categories then your enriched category can be thought of as having a natural ordinary category structure as well. However, identifying the ordinary category with the enriched category would be a slight abuse of notation, since the hom objects in the enriched category are objects of $V$ and the hom objects of the associated ordinary category are objects of $mathbf{Set}$. That's generally ok, since we usually speak of abelian groups and topological spaces as being sets.
However, this distinction between $V$-enriched categories and their associated ordinary categories (when $V$ is concrete) is very relevant here, because one should be careful about what we mean by an equivalence of categories.
If $C$ and $D$ are $V$-enriched categories, then an equivalence of those categories should be a pair of $V$-functors $F:Cto D$ and $G:Dto C$ with $Fcirc G simeq 1_D$ and $Gcirc F simeq 1_C$ (the natural isomorphisms here should also be $V$-natural isomorphisms). With this notion of equivalence, an equivalence between $C$ and $D$ would naturally preserve the enrichment, and this is the correct notion of an equivalence between $C$ and $D$.
As Eric Wofsey has shown with his example, not only is there no reason to think that an equivalence between the associated ordinary categories would induce an equivalence between the original $V$-enriched categories, it is not true. The main point of my answer is that there aren't even necessarily ordinary categories associated to enriched categories, since your hom objects don't necessarily even lie in a concrete category (Yes I know we can always take the generalized elements of the unit to get an ordinary category from an enriched category, but that's beside the point here).
answered Feb 1 at 23:56
jgonjgon
16.5k32143
$endgroup$
add a comment |
$begingroup$
Eric Wofsey has already given a great answer (+1) giving an example of two equivalent categories enriched over abelian groups where the abelian groups fail to be preserved and also added a good point about the preservation of enrichment in additive categories.
However, I'd like to add an additional point.
Enriched categories are not (necessarily) ordinary categories. In fact, I'd like to just say that they are not ordinary categories. If the enriching category, $V$, is concrete, as is the case with $mathbf{Top}$ or $mathbf{Ab}$ enriched categories then your enriched category can be thought of as having a natural ordinary category structure as well. However, identifying the ordinary category with the enriched category would be a slight abuse of notation, since the hom objects in the enriched category are objects of $V$ and the hom objects of the associated ordinary category are objects of $mathbf{Set}$. That's generally ok, since we usually speak of abelian groups and topological spaces as being sets.
However, this distinction between $V$-enriched categories and their associated ordinary categories (when $V$ is concrete) is very relevant here, because one should be careful about what we mean by an equivalence of categories.
If $C$ and $D$ are $V$-enriched categories, then an equivalence of those categories should be a pair of $V$-functors $F:Cto D$ and $G:Dto C$ with $Fcirc G simeq 1_D$ and $Gcirc F simeq 1_C$ (the natural isomorphisms here should also be $V$-natural isomorphisms). With this notion of equivalence, an equivalence between $C$ and $D$ would naturally preserve the enrichment, and this is the correct notion of an equivalence between $C$ and $D$.
As Eric Wofsey has shown with his example, not only is there no reason to think that an equivalence between the associated ordinary categories would induce an equivalence between the original $V$-enriched categories, it is not true. The main point of my answer is that there aren't even necessarily ordinary categories associated to enriched categories, since your hom objects don't necessarily even lie in a concrete category (Yes I know we can always take the generalized elements of the unit to get an ordinary category from an enriched category, but that's beside the point here).
answered Feb 1 at 23:56
jgonjgon
16.5k32143
$endgroup$
Eric Wofsey has already given a great answer (+1) giving an example of two equivalent categories enriched over abelian groups where the abelian groups fail to be preserved and also added a good point about the preservation of enrichment in additive categories.
However, I'd like to add an additional point.
Enriched categories are not (necessarily) ordinary categories. In fact, I'd like to just say that they are not ordinary categories. If the enriching category, $V$, is concrete, as is the case with $mathbf{Top}$ or $mathbf{Ab}$ enriched categories then your enriched category can be thought of as having a natural ordinary category structure as well. However, identifying the ordinary category with the enriched category would be a slight abuse of notation, since the hom objects in the enriched category are objects of $V$ and the hom objects of the associated ordinary category are objects of $mathbf{Set}$. That's generally ok, since we usually speak of abelian groups and topological spaces as being sets.
However, this distinction between $V$-enriched categories and their associated ordinary categories (when $V$ is concrete) is very relevant here, because one should be careful about what we mean by an equivalence of categories.
If $C$ and $D$ are $V$-enriched categories, then an equivalence of those categories should be a pair of $V$-functors $F:Cto D$ and $G:Dto C$ with $Fcirc G simeq 1_D$ and $Gcirc F simeq 1_C$ (the natural isomorphisms here should also be $V$-natural isomorphisms). With this notion of equivalence, an equivalence between $C$ and $D$ would naturally preserve the enrichment, and this is the correct notion of an equivalence between $C$ and $D$.
As Eric Wofsey has shown with his example, not only is there no reason to think that an equivalence between the associated ordinary categories would induce an equivalence between the original $V$-enriched categories, it is not true. The main point of my answer is that there aren't even necessarily ordinary categories associated to enriched categories, since your hom objects don't necessarily even lie in a concrete category (Yes I know we can always take the generalized elements of the unit to get an ordinary category from an enriched category, but that's beside the point here).
answered Feb 1 at 23:56
jgonjgon
16.5k32143
answered Feb 1 at 23:56
jgonjgon
16.5k32143
answered Feb 1 at 23:56
jgonjgon
16.5k32143
answered Feb 1 at 23:56
jgonjgon
16.5k32143
16.5k32143
add a comment |
add a comment |
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