Proof for limit of the identity functor Id ∷ C → C is the initial object
$begingroup$
I want to proove
Limit of the identity functor Id ∷ C → C is the initial object
I get that the limit object would be having morphisms going out to every other object. (Altogether forming the universal cone) So thats one condition for becoming an initial object.
But theres another condition to be fulfilled to be an initial object, that there must be only one morphism for each other object.
Being a limit object doesn't mean there's only one cone on it. There could be more natural transformations from apex functor to Id functor as long as there's factorizing morphism going into itself right?
How can I proove it?
category-theory
asked Jan 3 at 12:17
IngunIngun
133
$endgroup$
add a comment |
$begingroup$
I want to proove
Limit of the identity functor Id ∷ C → C is the initial object
I get that the limit object would be having morphisms going out to every other object. (Altogether forming the universal cone) So thats one condition for becoming an initial object.
But theres another condition to be fulfilled to be an initial object, that there must be only one morphism for each other object.
Being a limit object doesn't mean there's only one cone on it. There could be more natural transformations from apex functor to Id functor as long as there's factorizing morphism going into itself right?
How can I proove it?
category-theory
asked Jan 3 at 12:17
IngunIngun
133
$endgroup$
$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
1
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17
add a comment |
$begingroup$
I want to proove
Limit of the identity functor Id ∷ C → C is the initial object
I get that the limit object would be having morphisms going out to every other object. (Altogether forming the universal cone) So thats one condition for becoming an initial object.
But theres another condition to be fulfilled to be an initial object, that there must be only one morphism for each other object.
Being a limit object doesn't mean there's only one cone on it. There could be more natural transformations from apex functor to Id functor as long as there's factorizing morphism going into itself right?
How can I proove it?
category-theory
asked Jan 3 at 12:17
IngunIngun
133
$endgroup$
I want to proove
Limit of the identity functor Id ∷ C → C is the initial object
I get that the limit object would be having morphisms going out to every other object. (Altogether forming the universal cone) So thats one condition for becoming an initial object.
But theres another condition to be fulfilled to be an initial object, that there must be only one morphism for each other object.
Being a limit object doesn't mean there's only one cone on it. There could be more natural transformations from apex functor to Id functor as long as there's factorizing morphism going into itself right?
How can I proove it?
category-theory
category-theory
asked Jan 3 at 12:17
IngunIngun
133
asked Jan 3 at 12:17
IngunIngun
133
edited Jan 3 at 13:04
Ingun
asked Jan 3 at 12:17
IngunIngun
133
asked Jan 3 at 12:17
IngunIngun
133
asked Jan 3 at 12:17
IngunIngun
133
133
$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
1
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17
add a comment |
$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
1
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17
$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
1
1
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17
add a comment |
1 Answer
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Just for the purpose of having the question answered on this site:
Suppose the identity functor $text{Id}colon Cto C$ has a limit. In particular this is a cone over $text{Id}$: an object $L$ in $C$ and an arrow $pi_Xcolon Lto X$ for every object $X$ in $C$, such that for every arrow $fcolon Yto X$ in $C$, we have $pi_X = fcirc pi_Y$.
First we argue that $pi_L = text{id}_L$. Note that $pi_Lcolon Lto L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = pi_X$ in the equation above, we have $pi_X = pi_X circ pi_L$. But since the cone is terminal, $text{id}_L$ is the unique arrow from the cone to itself, and $pi_L = text{id}_L$.
Next we argue that for any object $X$ in $C$, $pi_X$ is the unique arrow $Lto X$. Indeed, for any arrow $fcolon Lto X$, taking $Y = L$ in the equation above, we have $pi_X = fcirc pi_L = fcirc text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
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add a comment |
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$begingroup$
Just for the purpose of having the question answered on this site:
Suppose the identity functor $text{Id}colon Cto C$ has a limit. In particular this is a cone over $text{Id}$: an object $L$ in $C$ and an arrow $pi_Xcolon Lto X$ for every object $X$ in $C$, such that for every arrow $fcolon Yto X$ in $C$, we have $pi_X = fcirc pi_Y$.
First we argue that $pi_L = text{id}_L$. Note that $pi_Lcolon Lto L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = pi_X$ in the equation above, we have $pi_X = pi_X circ pi_L$. But since the cone is terminal, $text{id}_L$ is the unique arrow from the cone to itself, and $pi_L = text{id}_L$.
Next we argue that for any object $X$ in $C$, $pi_X$ is the unique arrow $Lto X$. Indeed, for any arrow $fcolon Lto X$, taking $Y = L$ in the equation above, we have $pi_X = fcirc pi_L = fcirc text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
$endgroup$
add a comment |
$begingroup$
Just for the purpose of having the question answered on this site:
Suppose the identity functor $text{Id}colon Cto C$ has a limit. In particular this is a cone over $text{Id}$: an object $L$ in $C$ and an arrow $pi_Xcolon Lto X$ for every object $X$ in $C$, such that for every arrow $fcolon Yto X$ in $C$, we have $pi_X = fcirc pi_Y$.
First we argue that $pi_L = text{id}_L$. Note that $pi_Lcolon Lto L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = pi_X$ in the equation above, we have $pi_X = pi_X circ pi_L$. But since the cone is terminal, $text{id}_L$ is the unique arrow from the cone to itself, and $pi_L = text{id}_L$.
Next we argue that for any object $X$ in $C$, $pi_X$ is the unique arrow $Lto X$. Indeed, for any arrow $fcolon Lto X$, taking $Y = L$ in the equation above, we have $pi_X = fcirc pi_L = fcirc text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
$endgroup$
add a comment |
$begingroup$
Just for the purpose of having the question answered on this site:
Suppose the identity functor $text{Id}colon Cto C$ has a limit. In particular this is a cone over $text{Id}$: an object $L$ in $C$ and an arrow $pi_Xcolon Lto X$ for every object $X$ in $C$, such that for every arrow $fcolon Yto X$ in $C$, we have $pi_X = fcirc pi_Y$.
First we argue that $pi_L = text{id}_L$. Note that $pi_Lcolon Lto L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = pi_X$ in the equation above, we have $pi_X = pi_X circ pi_L$. But since the cone is terminal, $text{id}_L$ is the unique arrow from the cone to itself, and $pi_L = text{id}_L$.
Next we argue that for any object $X$ in $C$, $pi_X$ is the unique arrow $Lto X$. Indeed, for any arrow $fcolon Lto X$, taking $Y = L$ in the equation above, we have $pi_X = fcirc pi_L = fcirc text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
$endgroup$
Just for the purpose of having the question answered on this site:
Suppose the identity functor $text{Id}colon Cto C$ has a limit. In particular this is a cone over $text{Id}$: an object $L$ in $C$ and an arrow $pi_Xcolon Lto X$ for every object $X$ in $C$, such that for every arrow $fcolon Yto X$ in $C$, we have $pi_X = fcirc pi_Y$.
First we argue that $pi_L = text{id}_L$. Note that $pi_Lcolon Lto L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = pi_X$ in the equation above, we have $pi_X = pi_X circ pi_L$. But since the cone is terminal, $text{id}_L$ is the unique arrow from the cone to itself, and $pi_L = text{id}_L$.
Next we argue that for any object $X$ in $C$, $pi_X$ is the unique arrow $Lto X$. Indeed, for any arrow $fcolon Lto X$, taking $Y = L$ in the equation above, we have $pi_X = fcirc pi_L = fcirc text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
answered Jan 3 at 18:29
Alex KruckmanAlex Kruckman
27.1k22556
27.1k22556
add a comment |
add a comment |
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$begingroup$
Here's what I see. Please use MathJax.
$endgroup$
– Shaun
Jan 3 at 12:35
$begingroup$
Look at Theorem 4.3 in ncatlab.org/nlab/show/initial+object.
$endgroup$
– mfox
Jan 3 at 12:52
1
$begingroup$
@Shaun I'm new. Sorry for that. Just fixed. Thank you for the tip!
$endgroup$
– Ingun
Jan 3 at 13:05
$begingroup$
@mfox Actually Lemma 4.2 had the proof I wanted. thank you so much!
$endgroup$
– Ingun
Jan 3 at 14:17