Relationship between $ operatorname{Hom} (A, -)$ and $ operatorname{Hom}(-, B)$ functors?
$begingroup$
Wiki states that:
The pair of functors $ DeclareMathOperator{Hom}{Hom}Hom(A, –)$ and $ Hom(–, B)$ are related in a natural manner.
Than there is a commuting diagram, which I can't understand. Let's take one of its paths:
$ Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B') xrightarrow{Hom(h: A rightarrow A', B')} Hom(A', B')$
So, $Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B')$ maps ${ A rightarrow B }$ arrows to the ${ A rightarrow B' }$ arrows by a known way. Here things getting unclear to me: according to the path above, next morphism is $Hom(h: A rightarrow A', B')$. The problem is that it does not work with arrows of type ${ A rightarrow B' }$, it instead expects arrows of type ${ A leftarrow B' }$.
Hence, I got stuck.
I assume that this is some kind of common "abuse of notation".
I would like to know how to read abovementioned diagram and define proper natural transformation (or natural isomorphism?) between the pair of functors $Hom(A, –)$ and $Hom(–, B)$?
category-theory definition
edited Jan 6 at 12:28
Bernard
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
$endgroup$
|
show 1 more comment
$begingroup$
Wiki states that:
The pair of functors $ DeclareMathOperator{Hom}{Hom}Hom(A, –)$ and $ Hom(–, B)$ are related in a natural manner.
Than there is a commuting diagram, which I can't understand. Let's take one of its paths:
$ Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B') xrightarrow{Hom(h: A rightarrow A', B')} Hom(A', B')$
So, $Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B')$ maps ${ A rightarrow B }$ arrows to the ${ A rightarrow B' }$ arrows by a known way. Here things getting unclear to me: according to the path above, next morphism is $Hom(h: A rightarrow A', B')$. The problem is that it does not work with arrows of type ${ A rightarrow B' }$, it instead expects arrows of type ${ A leftarrow B' }$.
Hence, I got stuck.
I assume that this is some kind of common "abuse of notation".
I would like to know how to read abovementioned diagram and define proper natural transformation (or natural isomorphism?) between the pair of functors $Hom(A, –)$ and $Hom(–, B)$?
category-theory definition
edited Jan 6 at 12:28
Bernard
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
$endgroup$
$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
$begingroup$
@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
$begingroup$
No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
$begingroup$
@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
$begingroup$
The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28
|
show 1 more comment
$begingroup$
Wiki states that:
The pair of functors $ DeclareMathOperator{Hom}{Hom}Hom(A, –)$ and $ Hom(–, B)$ are related in a natural manner.
Than there is a commuting diagram, which I can't understand. Let's take one of its paths:
$ Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B') xrightarrow{Hom(h: A rightarrow A', B')} Hom(A', B')$
So, $Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B')$ maps ${ A rightarrow B }$ arrows to the ${ A rightarrow B' }$ arrows by a known way. Here things getting unclear to me: according to the path above, next morphism is $Hom(h: A rightarrow A', B')$. The problem is that it does not work with arrows of type ${ A rightarrow B' }$, it instead expects arrows of type ${ A leftarrow B' }$.
Hence, I got stuck.
I assume that this is some kind of common "abuse of notation".
I would like to know how to read abovementioned diagram and define proper natural transformation (or natural isomorphism?) between the pair of functors $Hom(A, –)$ and $Hom(–, B)$?
category-theory definition
edited Jan 6 at 12:28
Bernard
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
$endgroup$
Wiki states that:
The pair of functors $ DeclareMathOperator{Hom}{Hom}Hom(A, –)$ and $ Hom(–, B)$ are related in a natural manner.
Than there is a commuting diagram, which I can't understand. Let's take one of its paths:
$ Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B') xrightarrow{Hom(h: A rightarrow A', B')} Hom(A', B')$
So, $Hom(A, B) xrightarrow{Hom(A, f: B rightarrow B')} Hom(A, B')$ maps ${ A rightarrow B }$ arrows to the ${ A rightarrow B' }$ arrows by a known way. Here things getting unclear to me: according to the path above, next morphism is $Hom(h: A rightarrow A', B')$. The problem is that it does not work with arrows of type ${ A rightarrow B' }$, it instead expects arrows of type ${ A leftarrow B' }$.
Hence, I got stuck.
I assume that this is some kind of common "abuse of notation".
I would like to know how to read abovementioned diagram and define proper natural transformation (or natural isomorphism?) between the pair of functors $Hom(A, –)$ and $Hom(–, B)$?
category-theory definition
category-theory definition
edited Jan 6 at 12:28
Bernard
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
edited Jan 6 at 12:28
Bernard
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
edited Jan 6 at 12:28
Bernard
119k639112
edited Jan 6 at 12:28
Bernard
119k639112
edited Jan 6 at 12:28
Bernard
119k639112
119k639112
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
asked Jan 6 at 10:04
Sereja BogolubovSereja Bogolubov
590211
590211
$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
$begingroup$
@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
$begingroup$
No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
$begingroup$
@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
$begingroup$
The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28
|
show 1 more comment
$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
$begingroup$
@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
$begingroup$
No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
$begingroup$
@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
$begingroup$
The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28
$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
$begingroup$
@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
$begingroup$
@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
$begingroup$
No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
$begingroup$
No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
$begingroup$
@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
$begingroup$
@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
$begingroup$
The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28
$begingroup$
The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $tau_B:=mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ for every object $B$ then $tau$ is a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$.
If $rho_A:=mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$ for every object $A$ then $rho$ is a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$.
Let $mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$ and $tau_B=mathsf{Hom}(h,B)$ is prescribed by: $$umapsto ucirc h$$
Further $f$ denotes an arrow in homset $mathcal C(B,B')$ and $rho_B=mathsf{Hom}(A,f)$ is prescribed by: $$umapsto fcirc u$$
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
$endgroup$
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
|
show 3 more comments
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1 Answer
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$begingroup$
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $tau_B:=mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ for every object $B$ then $tau$ is a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$.
If $rho_A:=mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$ for every object $A$ then $rho$ is a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$.
Let $mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$ and $tau_B=mathsf{Hom}(h,B)$ is prescribed by: $$umapsto ucirc h$$
Further $f$ denotes an arrow in homset $mathcal C(B,B')$ and $rho_B=mathsf{Hom}(A,f)$ is prescribed by: $$umapsto fcirc u$$
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
$endgroup$
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
|
show 3 more comments
$begingroup$
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $tau_B:=mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ for every object $B$ then $tau$ is a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$.
If $rho_A:=mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$ for every object $A$ then $rho$ is a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$.
Let $mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$ and $tau_B=mathsf{Hom}(h,B)$ is prescribed by: $$umapsto ucirc h$$
Further $f$ denotes an arrow in homset $mathcal C(B,B')$ and $rho_B=mathsf{Hom}(A,f)$ is prescribed by: $$umapsto fcirc u$$
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
$endgroup$
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
|
show 3 more comments
$begingroup$
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $tau_B:=mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ for every object $B$ then $tau$ is a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$.
If $rho_A:=mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$ for every object $A$ then $rho$ is a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$.
Let $mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$ and $tau_B=mathsf{Hom}(h,B)$ is prescribed by: $$umapsto ucirc h$$
Further $f$ denotes an arrow in homset $mathcal C(B,B')$ and $rho_B=mathsf{Hom}(A,f)$ is prescribed by: $$umapsto fcirc u$$
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
$endgroup$
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $tau_B:=mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ for every object $B$ then $tau$ is a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$.
If $rho_A:=mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$ for every object $A$ then $rho$ is a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$.
Let $mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$ and $tau_B=mathsf{Hom}(h,B)$ is prescribed by: $$umapsto ucirc h$$
Further $f$ denotes an arrow in homset $mathcal C(B,B')$ and $rho_B=mathsf{Hom}(A,f)$ is prescribed by: $$umapsto fcirc u$$
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
answered Jan 6 at 11:03
drhabdrhab
98.9k544129
98.9k544129
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
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– Giorgio Mossa
Jan 6 at 14:27
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@GiorgioMossa, what is $Hom(B, h)$ then?
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– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
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– drhab
Jan 6 at 14:29
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A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
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– Giorgio Mossa
Jan 6 at 14:30
|
show 3 more comments
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I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
$begingroup$
I can't follow. AFAIU, $Hom(h, B)$ is $Hom(B, A) rightarrow Hom(B, A')$, but your statement is on the contrary. And what does it mean, that $mathcal C^{text{op}}(A,A')=mathcal C(A',A)$? Do you mean that both sets are isomorphic?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:22
1
1
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@SerejaBogolubov, you are mistaken, $Hom(h,B) colon Hom(A,B) to Hom(A',B)$. The equation $mathcal C^text{op}(A,A') = mathcal C(A,A')$ is the definition of the hom-set of arrows from $A$ to $A'$ into the opposite category $mathcal C^text{op}$: the category obtained by $mathcal C$ reversing the direction of the arrows.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:27
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
@GiorgioMossa, what is $Hom(B, h)$ then?
$endgroup$
– Sereja Bogolubov
Jan 6 at 14:28
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
Let $h:A'to A$. Then $mathsf{Hom}(h,B)$ is a function that has $mathsf{Hom}(A,B)$ as domain. Element $uinmathsf{Hom}(A,B)$ is sent to $ucirc hinmathsf{Hom}(A',B)$. Are you familiar with opposite category?
$endgroup$
– drhab
Jan 6 at 14:29
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
$begingroup$
A function of type $Hom(B,A') to Hom(B,A)$, assuming $h colon A to A'$.
$endgroup$
– Giorgio Mossa
Jan 6 at 14:30
|
show 3 more comments
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$begingroup$
If $mathcal C$ denotes the category then $mathsf{Hom}(A,-):mathcal Ctomathbf{Set}$ but $mathsf{Hom}(-,B):mathcal C^{text{op}}tomathbf{Set}$. So they have different domains so that there is no natural transformation.
$endgroup$
– drhab
Jan 6 at 10:10
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@drhab, do you claim that wiki is mistaken and $Hom(A, -)$, $Hom(-, B)$ do not relate in a natural way?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:14
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No I am not saying that. I am only saying that a natural transformation between both functors (that you would like to define) does not exist. For that you need functors $F,G:mathcal Ctomathcal C'$ having the same domain and codomain. So "related in a natural manner" is not the same as "existence of natural transformation".
$endgroup$
– drhab
Jan 6 at 10:17
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@drhab, why then wiki states that both relate in a natural way, linking natural transformation?
$endgroup$
– Sereja Bogolubov
Jan 6 at 10:21
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The diagram shown looks quite natural. It visualizes a natural transformation $mathsf{Hom}(A,-)tomathsf{Hom}(A',-)$ and also a natural transformation $mathsf{Hom}(-,B)tomathsf{Hom}(-,B')$. For the first we have components $mathsf{Hom}(h,B):mathsf{Hom}(A,B)tomathsf{Hom}(A',B)$ and for the second components $mathsf{Hom}(A,f):mathsf{Hom}(A,B)tomathsf{Hom}(A,B')$. In both cases the transformation is completely determined by an arrow ($h$ in first and $f$ in second case) corresponding with Yoneda.
$endgroup$
– drhab
Jan 6 at 10:28