Mixing
Can Endofunctors be seen as the definition of equivalence classes? How is this useful?
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Functions can be seen as identifying equivalence classes on a set. For instance, take a set $X$ and an endo-function $f: X rightarrow X$. Suppose the pre-image of $x_0 in X$ is a set $A subset X$, label this preimage with $x_0$. Take this for all elements $x in X$. Then we have a definition of equivalence classes labelled by elements of $X$. This applies to every function.
Next, consider an endofunctor, $F$, on a category $C$. We take the view of an endofunctor as mapping arrows to arrows. Every arrow has a pre-image and this pre-image is a a set of arrows. Thus, the endofunctor $F$ defines a set of equivalence classes over the arrows of $C$ and they are themselves labelled by arrows of $C$.
Is this a standard interpretation? Does it have a use?
category-theory
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
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add a comment |
$begingroup$
Functions can be seen as identifying equivalence classes on a set. For instance, take a set $X$ and an endo-function $f: X rightarrow X$. Suppose the pre-image of $x_0 in X$ is a set $A subset X$, label this preimage with $x_0$. Take this for all elements $x in X$. Then we have a definition of equivalence classes labelled by elements of $X$. This applies to every function.
Next, consider an endofunctor, $F$, on a category $C$. We take the view of an endofunctor as mapping arrows to arrows. Every arrow has a pre-image and this pre-image is a a set of arrows. Thus, the endofunctor $F$ defines a set of equivalence classes over the arrows of $C$ and they are themselves labelled by arrows of $C$.
Is this a standard interpretation? Does it have a use?
category-theory
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
$endgroup$
$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
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@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49
add a comment |
$begingroup$
Functions can be seen as identifying equivalence classes on a set. For instance, take a set $X$ and an endo-function $f: X rightarrow X$. Suppose the pre-image of $x_0 in X$ is a set $A subset X$, label this preimage with $x_0$. Take this for all elements $x in X$. Then we have a definition of equivalence classes labelled by elements of $X$. This applies to every function.
Next, consider an endofunctor, $F$, on a category $C$. We take the view of an endofunctor as mapping arrows to arrows. Every arrow has a pre-image and this pre-image is a a set of arrows. Thus, the endofunctor $F$ defines a set of equivalence classes over the arrows of $C$ and they are themselves labelled by arrows of $C$.
Is this a standard interpretation? Does it have a use?
category-theory
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
$endgroup$
Functions can be seen as identifying equivalence classes on a set. For instance, take a set $X$ and an endo-function $f: X rightarrow X$. Suppose the pre-image of $x_0 in X$ is a set $A subset X$, label this preimage with $x_0$. Take this for all elements $x in X$. Then we have a definition of equivalence classes labelled by elements of $X$. This applies to every function.
Next, consider an endofunctor, $F$, on a category $C$. We take the view of an endofunctor as mapping arrows to arrows. Every arrow has a pre-image and this pre-image is a a set of arrows. Thus, the endofunctor $F$ defines a set of equivalence classes over the arrows of $C$ and they are themselves labelled by arrows of $C$.
Is this a standard interpretation? Does it have a use?
category-theory
category-theory
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
asked Jan 31 at 16:54
Ben SprottBen Sprott
436312
436312
$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
$begingroup$
@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49
add a comment |
$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
$begingroup$
@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49
$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
$begingroup$
@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49
add a comment |
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$begingroup$
But you're just talking about a special case of the first idea, when $f$ happens to be the morphism part of a functor, as far as I can tell.
$endgroup$
– Kevin Carlson
Jan 31 at 18:02
$begingroup$
@KevinCarlson Yes, I think so. This view, of functions or functors, is it a general tool? Does it have a use?
$endgroup$
– Ben Sprott
Jan 31 at 19:05
$begingroup$
@KevinCarlson Yes, I see them as being the same thing. The use in Category Theory would be different though, if there is some.
$endgroup$
– Ben Sprott
Jan 31 at 19:10
$begingroup$
What you describe does not produce a partition because there are elements of $X$ whose preimage is empty. You could require $f$ to be surjective (or I guess you could just filter out the empty sets which would be equivalent to considering $f$ as a [surjective] function onto its image) or you could define an equivalence relation via $x_1sim x_2 iff f(x_1)=f(x_2)$ and then calculate equivalence classes in the usual way. Also, there's no reason $f$ needs to be an endofunction, and there's not a lot of point in "labeling" the equivalence classes. The labels won't be representatives, for example.
$endgroup$
– Derek Elkins
Jan 31 at 20:49