Showing when two functors are naturally isomorphic, if one is faithful, then the other also is. [duplicate]
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What properties do natural isomorphisms between functors preserve?
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Supposing we have a natural isomorphism $tau : S rightarrow T$ between functors $S,T : mathscr{C} rightarrow mathscr{D}$, how exactly do we show that if $S$ is faithful, then so is $T$?
If $S$ is faithful, then it is injective on the hom-sets, i.e. if $f = g$, then $S(f) = S(g)$. I also know that the components of the natural transformation form a commutative square defined by $tau circ S(f) = T(f) circ tau$, and that the ultimate implication I want to show is (probably along the lines of?)
$$f = g ;;; Rightarrow ;;; S(f)=S(g) ;;; Rightarrow ;;; T(f)=T(g)$$
(with the first implication being given as stated by $S$ being faithful).
I've tried playing with around with some of the compositions but the closest I've gotten would be $T(tau(f)) = T(tau(g))$ (which came from $S(f)=S(g)$, composing on the right by $tau$ which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure.
Anyhow, sorry for the dumb question, but any ideas?
category-theory functors natural-transformations
asked Nov 20 at 1:09
Eevee Trainer
1,474216
marked as duplicate by Arnaud D., jgon, Rushabh Mehta, amWhy, KReiser Nov 22 at 0:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
1
down vote
favorite
This question already has an answer here:
What properties do natural isomorphisms between functors preserve?
1 answer
Supposing we have a natural isomorphism $tau : S rightarrow T$ between functors $S,T : mathscr{C} rightarrow mathscr{D}$, how exactly do we show that if $S$ is faithful, then so is $T$?
If $S$ is faithful, then it is injective on the hom-sets, i.e. if $f = g$, then $S(f) = S(g)$. I also know that the components of the natural transformation form a commutative square defined by $tau circ S(f) = T(f) circ tau$, and that the ultimate implication I want to show is (probably along the lines of?)
$$f = g ;;; Rightarrow ;;; S(f)=S(g) ;;; Rightarrow ;;; T(f)=T(g)$$
(with the first implication being given as stated by $S$ being faithful).
I've tried playing with around with some of the compositions but the closest I've gotten would be $T(tau(f)) = T(tau(g))$ (which came from $S(f)=S(g)$, composing on the right by $tau$ which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure.
Anyhow, sorry for the dumb question, but any ideas?
category-theory functors natural-transformations
asked Nov 20 at 1:09
Eevee Trainer
1,474216
marked as duplicate by Arnaud D., jgon, Rushabh Mehta, amWhy, KReiser Nov 22 at 0:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
What properties do natural isomorphisms between functors preserve?
1 answer
Supposing we have a natural isomorphism $tau : S rightarrow T$ between functors $S,T : mathscr{C} rightarrow mathscr{D}$, how exactly do we show that if $S$ is faithful, then so is $T$?
If $S$ is faithful, then it is injective on the hom-sets, i.e. if $f = g$, then $S(f) = S(g)$. I also know that the components of the natural transformation form a commutative square defined by $tau circ S(f) = T(f) circ tau$, and that the ultimate implication I want to show is (probably along the lines of?)
$$f = g ;;; Rightarrow ;;; S(f)=S(g) ;;; Rightarrow ;;; T(f)=T(g)$$
(with the first implication being given as stated by $S$ being faithful).
I've tried playing with around with some of the compositions but the closest I've gotten would be $T(tau(f)) = T(tau(g))$ (which came from $S(f)=S(g)$, composing on the right by $tau$ which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure.
Anyhow, sorry for the dumb question, but any ideas?
category-theory functors natural-transformations
asked Nov 20 at 1:09
Eevee Trainer
1,474216
This question already has an answer here:
What properties do natural isomorphisms between functors preserve?
1 answer
Supposing we have a natural isomorphism $tau : S rightarrow T$ between functors $S,T : mathscr{C} rightarrow mathscr{D}$, how exactly do we show that if $S$ is faithful, then so is $T$?
If $S$ is faithful, then it is injective on the hom-sets, i.e. if $f = g$, then $S(f) = S(g)$. I also know that the components of the natural transformation form a commutative square defined by $tau circ S(f) = T(f) circ tau$, and that the ultimate implication I want to show is (probably along the lines of?)
$$f = g ;;; Rightarrow ;;; S(f)=S(g) ;;; Rightarrow ;;; T(f)=T(g)$$
(with the first implication being given as stated by $S$ being faithful).
I've tried playing with around with some of the compositions but the closest I've gotten would be $T(tau(f)) = T(tau(g))$ (which came from $S(f)=S(g)$, composing on the right by $tau$ which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure.
Anyhow, sorry for the dumb question, but any ideas?
This question already has an answer here:
What properties do natural isomorphisms between functors preserve?
1 answer
category-theory functors natural-transformations
category-theory functors natural-transformations
asked Nov 20 at 1:09
Eevee Trainer
1,474216
asked Nov 20 at 1:09
Eevee Trainer
1,474216
edited Nov 20 at 1:16
asked Nov 20 at 1:09
Eevee Trainer
1,474216
asked Nov 20 at 1:09
Eevee Trainer
1,474216
asked Nov 20 at 1:09
Eevee Trainer
1,474216
1,474216
marked as duplicate by Arnaud D., jgon, Rushabh Mehta, amWhy, KReiser Nov 22 at 0:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud D., jgon, Rushabh Mehta, amWhy, KReiser Nov 22 at 0:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47
add a comment |
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47
add a comment |
1 Answer
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up vote
2
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The point is that $T(f)=tau_y S(f)tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The point is that $T(f)=tau_y S(f)tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
add a comment |
up vote
2
down vote
accepted
The point is that $T(f)=tau_y S(f)tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The point is that $T(f)=tau_y S(f)tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
The point is that $T(f)=tau_y S(f)tau_x^{-1}$. So if $T(f)=T(g)$, then the same thing is true for $S(f)$ and $S(g)$, up to composition with some isomorphisms. Cancel the isomorphisms!
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
answered Nov 20 at 1:34
Kevin Carlson
32.1k23270
32.1k23270
add a comment |
add a comment |
Your implication regarding $S$ being faithful actually says that $S$ is a function.
– Kevin Carlson
Nov 20 at 1:32
Yeah you’re running implications the wrong way.
– Randall
Nov 20 at 1:33
@KevinCarlson (For the record I get your answer proper, I'm just curious.) Since $S$ being faithful means $S$'s arrow function is injective ... I'm not sure what the issue is exactly? In this context is $S$ not a function, because - again, in this context - we're looking at the behavior of $S$'s arrow function?
– Eevee Trainer
Nov 20 at 2:02
@Randall Thanks, I got the arrows mixed up after all. I think I usually approached it from the "contrapositive" view ($xneq y Rightarrow f(x) neq f(y)$) looking back at my older coursework and I guess I got the wires crossed since then.
– Eevee Trainer
Nov 20 at 2:04
The proof is the same as the one given here math.stackexchange.com/questions/3004250/…, except the compositions are the other way around -- duality!
– Musa Al-hassy
Nov 20 at 2:47