Kernels in a category with zero objects: essential uniqueness
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An object $X$ of a category $mathsf{C}$ is a zero object if it is both initial and terminal, that is, if for any $Y in mathsf{C}$ there are unique morphisms $Yto X$ and $Xto Y$.
We define a kernel of a morphism $fcolon Xto Y$ as an equalizer of $f$ and $g$, where $g$ is the unique morphism $X to 0 to Y$ for a zero object $0$.
However, what if a category has more than one zero object? I know that they are uniquely isomorphic, but how does it help? How can we speak of the essentialy uniqueness of a kernel?
It is known that two different objects $X$ and $Y$ satisfying the same universal properties are isomorphic. However, given a morphism $fcolon Xto Y$ and zero objects $0_a$ and $0_b$, being an equalizer of $f$ together with $Xto 0_a to Y$ and being an equalizer of $f$ together with $X to 0_b to Y$ are two different universal properties.
category-theory
edited Nov 22 at 11:33
Oskar
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
add a comment |
up vote
0
down vote
favorite
An object $X$ of a category $mathsf{C}$ is a zero object if it is both initial and terminal, that is, if for any $Y in mathsf{C}$ there are unique morphisms $Yto X$ and $Xto Y$.
We define a kernel of a morphism $fcolon Xto Y$ as an equalizer of $f$ and $g$, where $g$ is the unique morphism $X to 0 to Y$ for a zero object $0$.
However, what if a category has more than one zero object? I know that they are uniquely isomorphic, but how does it help? How can we speak of the essentialy uniqueness of a kernel?
It is known that two different objects $X$ and $Y$ satisfying the same universal properties are isomorphic. However, given a morphism $fcolon Xto Y$ and zero objects $0_a$ and $0_b$, being an equalizer of $f$ together with $Xto 0_a to Y$ and being an equalizer of $f$ together with $X to 0_b to Y$ are two different universal properties.
category-theory
edited Nov 22 at 11:33
Oskar
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
An object $X$ of a category $mathsf{C}$ is a zero object if it is both initial and terminal, that is, if for any $Y in mathsf{C}$ there are unique morphisms $Yto X$ and $Xto Y$.
We define a kernel of a morphism $fcolon Xto Y$ as an equalizer of $f$ and $g$, where $g$ is the unique morphism $X to 0 to Y$ for a zero object $0$.
However, what if a category has more than one zero object? I know that they are uniquely isomorphic, but how does it help? How can we speak of the essentialy uniqueness of a kernel?
It is known that two different objects $X$ and $Y$ satisfying the same universal properties are isomorphic. However, given a morphism $fcolon Xto Y$ and zero objects $0_a$ and $0_b$, being an equalizer of $f$ together with $Xto 0_a to Y$ and being an equalizer of $f$ together with $X to 0_b to Y$ are two different universal properties.
category-theory
edited Nov 22 at 11:33
Oskar
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
An object $X$ of a category $mathsf{C}$ is a zero object if it is both initial and terminal, that is, if for any $Y in mathsf{C}$ there are unique morphisms $Yto X$ and $Xto Y$.
We define a kernel of a morphism $fcolon Xto Y$ as an equalizer of $f$ and $g$, where $g$ is the unique morphism $X to 0 to Y$ for a zero object $0$.
However, what if a category has more than one zero object? I know that they are uniquely isomorphic, but how does it help? How can we speak of the essentialy uniqueness of a kernel?
It is known that two different objects $X$ and $Y$ satisfying the same universal properties are isomorphic. However, given a morphism $fcolon Xto Y$ and zero objects $0_a$ and $0_b$, being an equalizer of $f$ together with $Xto 0_a to Y$ and being an equalizer of $f$ together with $X to 0_b to Y$ are two different universal properties.
category-theory
category-theory
edited Nov 22 at 11:33
Oskar
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
edited Nov 22 at 11:33
Oskar
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
edited Nov 22 at 11:33
Oskar
2,7361718
edited Nov 22 at 11:33
Oskar
2,7361718
edited Nov 22 at 11:33
Oskar
2,7361718
2,7361718
asked Nov 22 at 11:20
Jxt921
938618
asked Nov 22 at 11:20
Jxt921
938618
asked Nov 22 at 11:20
Jxt921
938618
938618
There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28
add a comment |
There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28
There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28
There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The zero morphism $Xto Y$ is already unique:
There are unique arrow $0_ato 0_b$ and $0_bto 0_a$, inverses of each other.
Also, by uniqueness $Xto 0_b = Xto 0_ato 0_b$ and $0_bto Y=0_bto 0_ato Y$.
Putting these together, we see that the two compositions $Xto 0_ato Y$ and $Xto 0_bto Y$ are equal.
answered Nov 22 at 11:40
Berci
59.3k23672
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The zero morphism $Xto Y$ is already unique:
There are unique arrow $0_ato 0_b$ and $0_bto 0_a$, inverses of each other.
Also, by uniqueness $Xto 0_b = Xto 0_ato 0_b$ and $0_bto Y=0_bto 0_ato Y$.
Putting these together, we see that the two compositions $Xto 0_ato Y$ and $Xto 0_bto Y$ are equal.
answered Nov 22 at 11:40
Berci
59.3k23672
add a comment |
up vote
3
down vote
accepted
The zero morphism $Xto Y$ is already unique:
There are unique arrow $0_ato 0_b$ and $0_bto 0_a$, inverses of each other.
Also, by uniqueness $Xto 0_b = Xto 0_ato 0_b$ and $0_bto Y=0_bto 0_ato Y$.
Putting these together, we see that the two compositions $Xto 0_ato Y$ and $Xto 0_bto Y$ are equal.
answered Nov 22 at 11:40
Berci
59.3k23672
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The zero morphism $Xto Y$ is already unique:
There are unique arrow $0_ato 0_b$ and $0_bto 0_a$, inverses of each other.
Also, by uniqueness $Xto 0_b = Xto 0_ato 0_b$ and $0_bto Y=0_bto 0_ato Y$.
Putting these together, we see that the two compositions $Xto 0_ato Y$ and $Xto 0_bto Y$ are equal.
answered Nov 22 at 11:40
Berci
59.3k23672
The zero morphism $Xto Y$ is already unique:
There are unique arrow $0_ato 0_b$ and $0_bto 0_a$, inverses of each other.
Also, by uniqueness $Xto 0_b = Xto 0_ato 0_b$ and $0_bto Y=0_bto 0_ato Y$.
Putting these together, we see that the two compositions $Xto 0_ato Y$ and $Xto 0_bto Y$ are equal.
answered Nov 22 at 11:40
Berci
59.3k23672
answered Nov 22 at 11:40
Berci
59.3k23672
answered Nov 22 at 11:40
Berci
59.3k23672
answered Nov 22 at 11:40
Berci
59.3k23672
59.3k23672
add a comment |
add a comment |
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There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism.
– Christoph
Nov 22 at 11:28