Do binary equalizers imply all finite equalizers
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If a category $mathcalC$ has all binary equalizers, does it necessarily have all finite equalizers?
If we take $mathcalC=Sets$ then we can always take the intersection of two equalizers with its inclusion mapping and at worst end up with the empty set and its trivial inclusion mapping, and more generally any category with an initial object has all equalizers trivially, but it isn't clear to me that binary equalizers in a vacuum yield all finite equalizers.
category-theory
asked Aug 12 at 5:12
Alec Rhea
703515
add a comment |Â
up vote
1
down vote
favorite
If a category $mathcalC$ has all binary equalizers, does it necessarily have all finite equalizers?
If we take $mathcalC=Sets$ then we can always take the intersection of two equalizers with its inclusion mapping and at worst end up with the empty set and its trivial inclusion mapping, and more generally any category with an initial object has all equalizers trivially, but it isn't clear to me that binary equalizers in a vacuum yield all finite equalizers.
category-theory
asked Aug 12 at 5:12
Alec Rhea
703515
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If a category $mathcalC$ has all binary equalizers, does it necessarily have all finite equalizers?
If we take $mathcalC=Sets$ then we can always take the intersection of two equalizers with its inclusion mapping and at worst end up with the empty set and its trivial inclusion mapping, and more generally any category with an initial object has all equalizers trivially, but it isn't clear to me that binary equalizers in a vacuum yield all finite equalizers.
category-theory
asked Aug 12 at 5:12
Alec Rhea
703515
If a category $mathcalC$ has all binary equalizers, does it necessarily have all finite equalizers?
If we take $mathcalC=Sets$ then we can always take the intersection of two equalizers with its inclusion mapping and at worst end up with the empty set and its trivial inclusion mapping, and more generally any category with an initial object has all equalizers trivially, but it isn't clear to me that binary equalizers in a vacuum yield all finite equalizers.
category-theory
asked Aug 12 at 5:12
Alec Rhea
703515
asked Aug 12 at 5:12
Alec Rhea
703515
asked Aug 12 at 5:12
Alec Rhea
703515
asked Aug 12 at 5:12
Alec Rhea
703515
703515
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and
$k$-fold equalisers. Let $f_1,ldots,f_k+1$ be parallel arrows.
Let $g$ be an equaliser of $f_1,ldots,f_k$ and $h$ an
equaliser of $f_1circ g$ and $f_k+1circ g$. Then $gcirc h$ is
an equaliser of $f_1,ldots,f_k+1$.
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and
$k$-fold equalisers. Let $f_1,ldots,f_k+1$ be parallel arrows.
Let $g$ be an equaliser of $f_1,ldots,f_k$ and $h$ an
equaliser of $f_1circ g$ and $f_k+1circ g$. Then $gcirc h$ is
an equaliser of $f_1,ldots,f_k+1$.
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
add a comment |Â
up vote
1
down vote
accepted
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and
$k$-fold equalisers. Let $f_1,ldots,f_k+1$ be parallel arrows.
Let $g$ be an equaliser of $f_1,ldots,f_k$ and $h$ an
equaliser of $f_1circ g$ and $f_k+1circ g$. Then $gcirc h$ is
an equaliser of $f_1,ldots,f_k+1$.
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and
$k$-fold equalisers. Let $f_1,ldots,f_k+1$ be parallel arrows.
Let $g$ be an equaliser of $f_1,ldots,f_k$ and $h$ an
equaliser of $f_1circ g$ and $f_k+1circ g$. Then $gcirc h$ is
an equaliser of $f_1,ldots,f_k+1$.
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and
$k$-fold equalisers. Let $f_1,ldots,f_k+1$ be parallel arrows.
Let $g$ be an equaliser of $f_1,ldots,f_k$ and $h$ an
equaliser of $f_1circ g$ and $f_k+1circ g$. Then $gcirc h$ is
an equaliser of $f_1,ldots,f_k+1$.
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
edited Aug 12 at 7:00
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
answered Aug 12 at 6:42
Lord Shark the Unknown
86.7k952112
86.7k952112
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
add a comment |Â
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
Much appreciated, but I think you meant $f_1circ g$, $f_k+1circ g$ and $gcirc h$ correct? I always read $circ$ as 'after', so your current order of composition would be for a coequalizer unless I'm mistaken?
– Alec Rhea
Aug 12 at 6:57
1
1
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
Quite likely$ $!
– Lord Shark the Unknown
Aug 12 at 7:00
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2879990%2fdo-binary-equalizers-imply-all-finite-equalizers%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
