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Examples of contravariant functors
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$begingroup$
I understand the definition and usefulness of the notion of functor.
But I am worrying about the usefulness of the notion of a contravariant functor. Wikipedia writes:
There are many constructions in mathematics that would be functors but for the fact that they "turn morphisms around" and "reverse composition". We then define a contravariant functor [...]
But why do they "turn morphisms around", wouldn't it be easier to do the same without the inversion of morphisms and composition?
So I guess it would be beneficial for me to know some examples of naturally occuring contravariant functors. So let me ask: what are some constructions in mathematics that naturally occur as contravariant functors instead of covariant functor?
soft-question category-theory examples-counterexamples big-list
asked 5 hours ago
user7280899user7280899
902517
$endgroup$
add a comment |
$begingroup$
I understand the definition and usefulness of the notion of functor.
But I am worrying about the usefulness of the notion of a contravariant functor. Wikipedia writes:
There are many constructions in mathematics that would be functors but for the fact that they "turn morphisms around" and "reverse composition". We then define a contravariant functor [...]
But why do they "turn morphisms around", wouldn't it be easier to do the same without the inversion of morphisms and composition?
So I guess it would be beneficial for me to know some examples of naturally occuring contravariant functors. So let me ask: what are some constructions in mathematics that naturally occur as contravariant functors instead of covariant functor?
soft-question category-theory examples-counterexamples big-list
asked 5 hours ago
user7280899user7280899
902517
$endgroup$
add a comment |
$begingroup$
I understand the definition and usefulness of the notion of functor.
But I am worrying about the usefulness of the notion of a contravariant functor. Wikipedia writes:
There are many constructions in mathematics that would be functors but for the fact that they "turn morphisms around" and "reverse composition". We then define a contravariant functor [...]
But why do they "turn morphisms around", wouldn't it be easier to do the same without the inversion of morphisms and composition?
So I guess it would be beneficial for me to know some examples of naturally occuring contravariant functors. So let me ask: what are some constructions in mathematics that naturally occur as contravariant functors instead of covariant functor?
soft-question category-theory examples-counterexamples big-list
asked 5 hours ago
user7280899user7280899
902517
$endgroup$
I understand the definition and usefulness of the notion of functor.
But I am worrying about the usefulness of the notion of a contravariant functor. Wikipedia writes:
There are many constructions in mathematics that would be functors but for the fact that they "turn morphisms around" and "reverse composition". We then define a contravariant functor [...]
But why do they "turn morphisms around", wouldn't it be easier to do the same without the inversion of morphisms and composition?
So I guess it would be beneficial for me to know some examples of naturally occuring contravariant functors. So let me ask: what are some constructions in mathematics that naturally occur as contravariant functors instead of covariant functor?
soft-question category-theory examples-counterexamples big-list
soft-question category-theory examples-counterexamples big-list
asked 5 hours ago
user7280899user7280899
902517
asked 5 hours ago
user7280899user7280899
902517
asked 5 hours ago
user7280899user7280899
902517
asked 5 hours ago
user7280899user7280899
902517
asked 5 hours ago
user7280899user7280899
902517
902517
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In linear algebra, taking the dual of a vector space is a contravariant functor from the category of vector spaces to itself : given a linear map $f:Vto W$, you get an induced map $f^*:W^*to V^*:varphimapsto varphicirc L$ between the dual spaces, and you can easily check that $(id_V)^*=id_{V^*}$ and $(gcirc f)^*=f^*circ g^*$ always hold. In fact this is the first example of functor that appears in Eilenberg and MacLane's original paper!
In fact this is a particular case of a general construction : given any category $mathcal{C}$, every object $X$ defines a functor $operatorname{Hom}(_,X):mathcal{C}to mathbf{Set}$ that takes an object $Y$ to $operatorname{Hom}(Y,X)$ and a morphism $f:Yto Z$ to the function
$$f^*:operatorname{Hom}(Z,X)to operatorname{Hom}(Y,X):gmapsto gcirc f.$$
Dual vector spaces correspond to the case where $mathcal{C}$ is the category of vector spaces over some field $k$ and $X=k$.
Moreover, in this answer I gave the contravariant powerset functor as another example; this is not quite of the form described above, but almost. In fact the correspondence between subsets of a set $Y$ and their characteristic functions defines a natural isomorphism between the contravariant functor $operatorname{Hom}(_,{0,1})$ and the contravariant powerset functor.
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
$endgroup$
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
add a comment |
$begingroup$
Let $R$ be a ring and $M$ be a left $R$-module. Then the functor
$$
F={rm Hom}_R(cdot, M)
$$
is a contravariant functor from the category of $R$-modules to the category of abelian groups. Switching sides, the functor ${rm Hom}_R(M,cdot)$ is covariant.
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
$endgroup$
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In linear algebra, taking the dual of a vector space is a contravariant functor from the category of vector spaces to itself : given a linear map $f:Vto W$, you get an induced map $f^*:W^*to V^*:varphimapsto varphicirc L$ between the dual spaces, and you can easily check that $(id_V)^*=id_{V^*}$ and $(gcirc f)^*=f^*circ g^*$ always hold. In fact this is the first example of functor that appears in Eilenberg and MacLane's original paper!
In fact this is a particular case of a general construction : given any category $mathcal{C}$, every object $X$ defines a functor $operatorname{Hom}(_,X):mathcal{C}to mathbf{Set}$ that takes an object $Y$ to $operatorname{Hom}(Y,X)$ and a morphism $f:Yto Z$ to the function
$$f^*:operatorname{Hom}(Z,X)to operatorname{Hom}(Y,X):gmapsto gcirc f.$$
Dual vector spaces correspond to the case where $mathcal{C}$ is the category of vector spaces over some field $k$ and $X=k$.
Moreover, in this answer I gave the contravariant powerset functor as another example; this is not quite of the form described above, but almost. In fact the correspondence between subsets of a set $Y$ and their characteristic functions defines a natural isomorphism between the contravariant functor $operatorname{Hom}(_,{0,1})$ and the contravariant powerset functor.
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
$endgroup$
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
add a comment |
$begingroup$
In linear algebra, taking the dual of a vector space is a contravariant functor from the category of vector spaces to itself : given a linear map $f:Vto W$, you get an induced map $f^*:W^*to V^*:varphimapsto varphicirc L$ between the dual spaces, and you can easily check that $(id_V)^*=id_{V^*}$ and $(gcirc f)^*=f^*circ g^*$ always hold. In fact this is the first example of functor that appears in Eilenberg and MacLane's original paper!
In fact this is a particular case of a general construction : given any category $mathcal{C}$, every object $X$ defines a functor $operatorname{Hom}(_,X):mathcal{C}to mathbf{Set}$ that takes an object $Y$ to $operatorname{Hom}(Y,X)$ and a morphism $f:Yto Z$ to the function
$$f^*:operatorname{Hom}(Z,X)to operatorname{Hom}(Y,X):gmapsto gcirc f.$$
Dual vector spaces correspond to the case where $mathcal{C}$ is the category of vector spaces over some field $k$ and $X=k$.
Moreover, in this answer I gave the contravariant powerset functor as another example; this is not quite of the form described above, but almost. In fact the correspondence between subsets of a set $Y$ and their characteristic functions defines a natural isomorphism between the contravariant functor $operatorname{Hom}(_,{0,1})$ and the contravariant powerset functor.
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
$endgroup$
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
add a comment |
$begingroup$
In linear algebra, taking the dual of a vector space is a contravariant functor from the category of vector spaces to itself : given a linear map $f:Vto W$, you get an induced map $f^*:W^*to V^*:varphimapsto varphicirc L$ between the dual spaces, and you can easily check that $(id_V)^*=id_{V^*}$ and $(gcirc f)^*=f^*circ g^*$ always hold. In fact this is the first example of functor that appears in Eilenberg and MacLane's original paper!
In fact this is a particular case of a general construction : given any category $mathcal{C}$, every object $X$ defines a functor $operatorname{Hom}(_,X):mathcal{C}to mathbf{Set}$ that takes an object $Y$ to $operatorname{Hom}(Y,X)$ and a morphism $f:Yto Z$ to the function
$$f^*:operatorname{Hom}(Z,X)to operatorname{Hom}(Y,X):gmapsto gcirc f.$$
Dual vector spaces correspond to the case where $mathcal{C}$ is the category of vector spaces over some field $k$ and $X=k$.
Moreover, in this answer I gave the contravariant powerset functor as another example; this is not quite of the form described above, but almost. In fact the correspondence between subsets of a set $Y$ and their characteristic functions defines a natural isomorphism between the contravariant functor $operatorname{Hom}(_,{0,1})$ and the contravariant powerset functor.
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
$endgroup$
In linear algebra, taking the dual of a vector space is a contravariant functor from the category of vector spaces to itself : given a linear map $f:Vto W$, you get an induced map $f^*:W^*to V^*:varphimapsto varphicirc L$ between the dual spaces, and you can easily check that $(id_V)^*=id_{V^*}$ and $(gcirc f)^*=f^*circ g^*$ always hold. In fact this is the first example of functor that appears in Eilenberg and MacLane's original paper!
In fact this is a particular case of a general construction : given any category $mathcal{C}$, every object $X$ defines a functor $operatorname{Hom}(_,X):mathcal{C}to mathbf{Set}$ that takes an object $Y$ to $operatorname{Hom}(Y,X)$ and a morphism $f:Yto Z$ to the function
$$f^*:operatorname{Hom}(Z,X)to operatorname{Hom}(Y,X):gmapsto gcirc f.$$
Dual vector spaces correspond to the case where $mathcal{C}$ is the category of vector spaces over some field $k$ and $X=k$.
Moreover, in this answer I gave the contravariant powerset functor as another example; this is not quite of the form described above, but almost. In fact the correspondence between subsets of a set $Y$ and their characteristic functions defines a natural isomorphism between the contravariant functor $operatorname{Hom}(_,{0,1})$ and the contravariant powerset functor.
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
answered 4 hours ago
Arnaud D.Arnaud D.
16.4k52445
16.4k52445
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
add a comment |
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
1
1
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
$begingroup$
As a small addition to this answer, it is sometimes also interesting to consider situations where $X$ is not an element of $mathcal C$ itself, but of a category of which $mathcal C$ is a (full) subcategory; for instance, when $mathcal C$ consists of compact Hausdorff spaces and $X = mathbb R$.
$endgroup$
– Mees de Vries
4 hours ago
add a comment |
$begingroup$
Let $R$ be a ring and $M$ be a left $R$-module. Then the functor
$$
F={rm Hom}_R(cdot, M)
$$
is a contravariant functor from the category of $R$-modules to the category of abelian groups. Switching sides, the functor ${rm Hom}_R(M,cdot)$ is covariant.
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
$endgroup$
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
add a comment |
$begingroup$
Let $R$ be a ring and $M$ be a left $R$-module. Then the functor
$$
F={rm Hom}_R(cdot, M)
$$
is a contravariant functor from the category of $R$-modules to the category of abelian groups. Switching sides, the functor ${rm Hom}_R(M,cdot)$ is covariant.
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
$endgroup$
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
add a comment |
$begingroup$
Let $R$ be a ring and $M$ be a left $R$-module. Then the functor
$$
F={rm Hom}_R(cdot, M)
$$
is a contravariant functor from the category of $R$-modules to the category of abelian groups. Switching sides, the functor ${rm Hom}_R(M,cdot)$ is covariant.
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
$endgroup$
Let $R$ be a ring and $M$ be a left $R$-module. Then the functor
$$
F={rm Hom}_R(cdot, M)
$$
is a contravariant functor from the category of $R$-modules to the category of abelian groups. Switching sides, the functor ${rm Hom}_R(M,cdot)$ is covariant.
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
answered 5 hours ago
Dietrich BurdeDietrich Burde
82.9k649107
82.9k649107
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
add a comment |
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
$begingroup$
I unfortunately don't see why this should be something that is naturally occuring.
$endgroup$
– user7280899
4 hours ago
add a comment |
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