How many morphisms from 1 to 1+1 can there be? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Can the inner structure of an object be systematically deduced from its position in the category?Terminology: Is there a name for a category with biproducts?Finiteness and cardinality in abstract categoriesExpressive power of first-order category theoryAn isomorphism of categoriesGeneralizing indexed coproduct from $mathrmSet$ to other monoidal categories“Maybe Monad” for multi-pointed objects?How many elementary embeddings can there be?Quotient-free monoidal categoriesConnectedness in extensive categories in terms of set of connected components
How many morphisms from 1 to 1+1 can there be?
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Can the inner structure of an object be systematically deduced from its position in the category?Terminology: Is there a name for a category with biproducts?Finiteness and cardinality in abstract categoriesExpressive power of first-order category theoryAn isomorphism of categoriesGeneralizing indexed coproduct from $mathrmSet$ to other monoidal categories“Maybe Monad” for multi-pointed objects?How many elementary embeddings can there be?Quotient-free monoidal categoriesConnectedness in extensive categories in terms of set of connected components
$begingroup$
Here is an interesting question raised by Alice Rhyl.
Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?
There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.
Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.
There can be four. Take $C = mathrmSet^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.
Indeed, any power of two is possible; just take $C = mathrmSet^n$.
What other numbers are possible? (I find finite cardinals more interesting here.)
So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?
(All the examples I've given above are categories of this sort.)
ct.category-theory
asked 4 hours ago
John BaezJohn Baez
8,7804699
$endgroup$
add a comment |
$begingroup$
Here is an interesting question raised by Alice Rhyl.
Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?
There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.
Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.
There can be four. Take $C = mathrmSet^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.
Indeed, any power of two is possible; just take $C = mathrmSet^n$.
What other numbers are possible? (I find finite cardinals more interesting here.)
So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?
(All the examples I've given above are categories of this sort.)
ct.category-theory
asked 4 hours ago
John BaezJohn Baez
8,7804699
$endgroup$
1
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
$begingroup$
Here is an interesting question raised by Alice Rhyl.
Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?
There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.
Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.
There can be four. Take $C = mathrmSet^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.
Indeed, any power of two is possible; just take $C = mathrmSet^n$.
What other numbers are possible? (I find finite cardinals more interesting here.)
So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?
(All the examples I've given above are categories of this sort.)
ct.category-theory
asked 4 hours ago
John BaezJohn Baez
8,7804699
$endgroup$
Here is an interesting question raised by Alice Rhyl.
Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?
There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.
Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.
There can be four. Take $C = mathrmSet^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.
Indeed, any power of two is possible; just take $C = mathrmSet^n$.
What other numbers are possible? (I find finite cardinals more interesting here.)
So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?
(All the examples I've given above are categories of this sort.)
ct.category-theory
ct.category-theory
asked 4 hours ago
John BaezJohn Baez
8,7804699
asked 4 hours ago
John BaezJohn Baez
8,7804699
edited 4 hours ago
John Baez
asked 4 hours ago
John BaezJohn Baez
8,7804699
asked 4 hours ago
John BaezJohn Baez
8,7804699
asked 4 hours ago
John BaezJohn Baez
8,7804699
8,7804699
1
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
1
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago
1
1
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328468%2fhow-many-morphisms-from-1-to-11-can-there-be%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
$endgroup$
add a comment |
$begingroup$
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
$endgroup$
add a comment |
$begingroup$
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
$endgroup$
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
edited 2 hours ago
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
answered 2 hours ago
Emil JeřábekEmil Jeřábek
30.5k389143
30.5k389143
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328468%2fhow-many-morphisms-from-1-to-11-can-there-be%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
If you restrict to extensive categories, then the set of morphism from $1$ to $1 coprod 1$ form a boolean algebra. Conversely, if $B$ is a boolean algebra, then in the topos of sheaves over its stone space, the morphisms from $1$ to $1 coprod 1$ corresponds exactly to elements of $B$. So if you only care about finite set you get exactly power of $2$ in this case.
$endgroup$
– Simon Henry
1 hour ago