Morphisms and objects of the $mathbb{SET}$?
$begingroup$
Whenever there is a morphism $f : A mapsto B$ and $g : B mapsto C$, there must exist $A mapsto C$ = $g circ f$. That's a reasonable requirement.
However, I am a programmer and mostly deal with $mathbb{SET}$: category of sets and single-variable functions between them.
It seems to be special. I'll give you an example: let's fix some $X in Obj(mathbb{SET})$. If I am not mistaken, $forall Y in Obj(mathbb{SET}): forall y in Y: exists f: X mapsto Y$ defined as a constant function that maps entire $X$ to a particular $y in Y$. It works in exactly the same way as a constant functor to a fixed object of a target category.
Now the tricky part: there is a set of functions ${ X mapsto Y }$ for all $X, Y$ and each such set is an object of the $mathbb{SET}$. Hence, there exists a morphism $m: X rightarrow Y$ and there exists yet another morphism $M : X rightarrow { X mapsto Y }$. Now, assume there also exists a morphism $k: Y rightarrow Y'$. It would be reasonable to immidiately reuqire existence of $M' : M rightarrow { X mapsto Y' }$ defined as $X rightarrow Y rightarrow Y'$ path in the $mathbb{SET}$. So end up with:
$X rightarrow { X mapsto Y } rightarrow { X mapsto Y' }$, which, by the requirement I've started from, yields an $X rightarrow { X mapsto Y' }$ moprhism.
$mathbb{SET}$ seems to be very special. When some $x in X$ is lifted to a $X mapsto Y$ in the abovementioned fashion, you kind of appear at two different places at the same time: on the one hand, you get a particular element of an appropariate ${ X mapsto Y }$; on the other hand - lifting takes you to a ${ X mapsto Y }$ - an object of the $mathbb{SET}$.
So my quesiton is kind of soft - doest it blur razor-sharp edges between morphisms and objects within the same category? Is it really that special or I am I just overreacting?
soft-question category-theory
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
$endgroup$
add a comment |
$begingroup$
Whenever there is a morphism $f : A mapsto B$ and $g : B mapsto C$, there must exist $A mapsto C$ = $g circ f$. That's a reasonable requirement.
However, I am a programmer and mostly deal with $mathbb{SET}$: category of sets and single-variable functions between them.
It seems to be special. I'll give you an example: let's fix some $X in Obj(mathbb{SET})$. If I am not mistaken, $forall Y in Obj(mathbb{SET}): forall y in Y: exists f: X mapsto Y$ defined as a constant function that maps entire $X$ to a particular $y in Y$. It works in exactly the same way as a constant functor to a fixed object of a target category.
Now the tricky part: there is a set of functions ${ X mapsto Y }$ for all $X, Y$ and each such set is an object of the $mathbb{SET}$. Hence, there exists a morphism $m: X rightarrow Y$ and there exists yet another morphism $M : X rightarrow { X mapsto Y }$. Now, assume there also exists a morphism $k: Y rightarrow Y'$. It would be reasonable to immidiately reuqire existence of $M' : M rightarrow { X mapsto Y' }$ defined as $X rightarrow Y rightarrow Y'$ path in the $mathbb{SET}$. So end up with:
$X rightarrow { X mapsto Y } rightarrow { X mapsto Y' }$, which, by the requirement I've started from, yields an $X rightarrow { X mapsto Y' }$ moprhism.
$mathbb{SET}$ seems to be very special. When some $x in X$ is lifted to a $X mapsto Y$ in the abovementioned fashion, you kind of appear at two different places at the same time: on the one hand, you get a particular element of an appropariate ${ X mapsto Y }$; on the other hand - lifting takes you to a ${ X mapsto Y }$ - an object of the $mathbb{SET}$.
So my quesiton is kind of soft - doest it blur razor-sharp edges between morphisms and objects within the same category? Is it really that special or I am I just overreacting?
soft-question category-theory
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
$endgroup$
1
$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
7
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08
add a comment |
$begingroup$
Whenever there is a morphism $f : A mapsto B$ and $g : B mapsto C$, there must exist $A mapsto C$ = $g circ f$. That's a reasonable requirement.
However, I am a programmer and mostly deal with $mathbb{SET}$: category of sets and single-variable functions between them.
It seems to be special. I'll give you an example: let's fix some $X in Obj(mathbb{SET})$. If I am not mistaken, $forall Y in Obj(mathbb{SET}): forall y in Y: exists f: X mapsto Y$ defined as a constant function that maps entire $X$ to a particular $y in Y$. It works in exactly the same way as a constant functor to a fixed object of a target category.
Now the tricky part: there is a set of functions ${ X mapsto Y }$ for all $X, Y$ and each such set is an object of the $mathbb{SET}$. Hence, there exists a morphism $m: X rightarrow Y$ and there exists yet another morphism $M : X rightarrow { X mapsto Y }$. Now, assume there also exists a morphism $k: Y rightarrow Y'$. It would be reasonable to immidiately reuqire existence of $M' : M rightarrow { X mapsto Y' }$ defined as $X rightarrow Y rightarrow Y'$ path in the $mathbb{SET}$. So end up with:
$X rightarrow { X mapsto Y } rightarrow { X mapsto Y' }$, which, by the requirement I've started from, yields an $X rightarrow { X mapsto Y' }$ moprhism.
$mathbb{SET}$ seems to be very special. When some $x in X$ is lifted to a $X mapsto Y$ in the abovementioned fashion, you kind of appear at two different places at the same time: on the one hand, you get a particular element of an appropariate ${ X mapsto Y }$; on the other hand - lifting takes you to a ${ X mapsto Y }$ - an object of the $mathbb{SET}$.
So my quesiton is kind of soft - doest it blur razor-sharp edges between morphisms and objects within the same category? Is it really that special or I am I just overreacting?
soft-question category-theory
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
$endgroup$
Whenever there is a morphism $f : A mapsto B$ and $g : B mapsto C$, there must exist $A mapsto C$ = $g circ f$. That's a reasonable requirement.
However, I am a programmer and mostly deal with $mathbb{SET}$: category of sets and single-variable functions between them.
It seems to be special. I'll give you an example: let's fix some $X in Obj(mathbb{SET})$. If I am not mistaken, $forall Y in Obj(mathbb{SET}): forall y in Y: exists f: X mapsto Y$ defined as a constant function that maps entire $X$ to a particular $y in Y$. It works in exactly the same way as a constant functor to a fixed object of a target category.
Now the tricky part: there is a set of functions ${ X mapsto Y }$ for all $X, Y$ and each such set is an object of the $mathbb{SET}$. Hence, there exists a morphism $m: X rightarrow Y$ and there exists yet another morphism $M : X rightarrow { X mapsto Y }$. Now, assume there also exists a morphism $k: Y rightarrow Y'$. It would be reasonable to immidiately reuqire existence of $M' : M rightarrow { X mapsto Y' }$ defined as $X rightarrow Y rightarrow Y'$ path in the $mathbb{SET}$. So end up with:
$X rightarrow { X mapsto Y } rightarrow { X mapsto Y' }$, which, by the requirement I've started from, yields an $X rightarrow { X mapsto Y' }$ moprhism.
$mathbb{SET}$ seems to be very special. When some $x in X$ is lifted to a $X mapsto Y$ in the abovementioned fashion, you kind of appear at two different places at the same time: on the one hand, you get a particular element of an appropariate ${ X mapsto Y }$; on the other hand - lifting takes you to a ${ X mapsto Y }$ - an object of the $mathbb{SET}$.
So my quesiton is kind of soft - doest it blur razor-sharp edges between morphisms and objects within the same category? Is it really that special or I am I just overreacting?
soft-question category-theory
soft-question category-theory
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
asked Dec 22 '18 at 16:55
Sereja BogolubovSereja Bogolubov
610211
610211
1
$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
7
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08
add a comment |
1
$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
7
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08
1
1
$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
7
7
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08
add a comment |
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$begingroup$
The only thing special about $Set$ is that $Hom(X,Y)$ is again a set for any two sets $X,Y$. Which is not ultimately special, the same happens for example in vector spaces. What you've discovered is functoriality of $Hom$.
$endgroup$
– freakish
Dec 22 '18 at 17:06
7
$begingroup$
It looks like you're noticing mainly that $mathbf{Set}$ is cartesian closed--i.e. you can curry functions whose domains are finite products.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 18:08