Mixing
Uniqueness of categorical products
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I'm currently studying category theory out of Awodey's excellent online book, but I'm having trouble seeing that the mapping between two product objects is necessarily an iso.
If $P,p_0:Prightarrow A,p_1:Prightarrow B$ and $X,x_0:Xrightarrow A,x_1:Xrightarrow B$ are product diagrams for $A$ and $B$ in a category $mathcalC$, then we obtain unique arrows $u_0:Xrightarrow P$ and $u_1:Prightarrow X$ which yield the identities $$x_0=p_0circ u_0,$$ $$p_0=x_0circ u_1,$$ which we can then substitute into each other to obtain $$x_0=x_0circ u_1circ u_0,$$ $$p_0=p_0circ u_0circ u_1.$$ At this stage the author asserts that these identities, together with the identities $$x_0=x_0circ1_X,$$ $$p_0=p_0circ1_P$$ and the uniqueness condition on $u_0$ and $u_1$ yield the desired equalities $u_0circ u_1=1_P$ and $u_1circ u_0=1_X$, but I can't quite see it.
If there is some other arrow $u_2:Prightarrow X$ such that $u_0circ u_2=1_P$ then I agree we would have a contradiction, since we then have $$p_0=p_0circ1_P=p_0circ u_0circ u_2=x_0circ u_2$$ and $u_1$ is unique with this property, but is the existence of $u_2$ necessary if $u_0circ u_1neq1_P$? Could the identity not hold and there be no such $u_2$?
(There are of course corresponding identities for $p_1$ and $x_1$, but they lead me to the same question.)
category-theory products
asked Jul 14 at 18:16
Alec Rhea
703515
add a comment |Â
up vote
4
down vote
favorite
I'm currently studying category theory out of Awodey's excellent online book, but I'm having trouble seeing that the mapping between two product objects is necessarily an iso.
If $P,p_0:Prightarrow A,p_1:Prightarrow B$ and $X,x_0:Xrightarrow A,x_1:Xrightarrow B$ are product diagrams for $A$ and $B$ in a category $mathcalC$, then we obtain unique arrows $u_0:Xrightarrow P$ and $u_1:Prightarrow X$ which yield the identities $$x_0=p_0circ u_0,$$ $$p_0=x_0circ u_1,$$ which we can then substitute into each other to obtain $$x_0=x_0circ u_1circ u_0,$$ $$p_0=p_0circ u_0circ u_1.$$ At this stage the author asserts that these identities, together with the identities $$x_0=x_0circ1_X,$$ $$p_0=p_0circ1_P$$ and the uniqueness condition on $u_0$ and $u_1$ yield the desired equalities $u_0circ u_1=1_P$ and $u_1circ u_0=1_X$, but I can't quite see it.
If there is some other arrow $u_2:Prightarrow X$ such that $u_0circ u_2=1_P$ then I agree we would have a contradiction, since we then have $$p_0=p_0circ1_P=p_0circ u_0circ u_2=x_0circ u_2$$ and $u_1$ is unique with this property, but is the existence of $u_2$ necessary if $u_0circ u_1neq1_P$? Could the identity not hold and there be no such $u_2$?
(There are of course corresponding identities for $p_1$ and $x_1$, but they lead me to the same question.)
category-theory products
asked Jul 14 at 18:16
Alec Rhea
703515
1
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm currently studying category theory out of Awodey's excellent online book, but I'm having trouble seeing that the mapping between two product objects is necessarily an iso.
If $P,p_0:Prightarrow A,p_1:Prightarrow B$ and $X,x_0:Xrightarrow A,x_1:Xrightarrow B$ are product diagrams for $A$ and $B$ in a category $mathcalC$, then we obtain unique arrows $u_0:Xrightarrow P$ and $u_1:Prightarrow X$ which yield the identities $$x_0=p_0circ u_0,$$ $$p_0=x_0circ u_1,$$ which we can then substitute into each other to obtain $$x_0=x_0circ u_1circ u_0,$$ $$p_0=p_0circ u_0circ u_1.$$ At this stage the author asserts that these identities, together with the identities $$x_0=x_0circ1_X,$$ $$p_0=p_0circ1_P$$ and the uniqueness condition on $u_0$ and $u_1$ yield the desired equalities $u_0circ u_1=1_P$ and $u_1circ u_0=1_X$, but I can't quite see it.
If there is some other arrow $u_2:Prightarrow X$ such that $u_0circ u_2=1_P$ then I agree we would have a contradiction, since we then have $$p_0=p_0circ1_P=p_0circ u_0circ u_2=x_0circ u_2$$ and $u_1$ is unique with this property, but is the existence of $u_2$ necessary if $u_0circ u_1neq1_P$? Could the identity not hold and there be no such $u_2$?
(There are of course corresponding identities for $p_1$ and $x_1$, but they lead me to the same question.)
category-theory products
asked Jul 14 at 18:16
Alec Rhea
703515
I'm currently studying category theory out of Awodey's excellent online book, but I'm having trouble seeing that the mapping between two product objects is necessarily an iso.
If $P,p_0:Prightarrow A,p_1:Prightarrow B$ and $X,x_0:Xrightarrow A,x_1:Xrightarrow B$ are product diagrams for $A$ and $B$ in a category $mathcalC$, then we obtain unique arrows $u_0:Xrightarrow P$ and $u_1:Prightarrow X$ which yield the identities $$x_0=p_0circ u_0,$$ $$p_0=x_0circ u_1,$$ which we can then substitute into each other to obtain $$x_0=x_0circ u_1circ u_0,$$ $$p_0=p_0circ u_0circ u_1.$$ At this stage the author asserts that these identities, together with the identities $$x_0=x_0circ1_X,$$ $$p_0=p_0circ1_P$$ and the uniqueness condition on $u_0$ and $u_1$ yield the desired equalities $u_0circ u_1=1_P$ and $u_1circ u_0=1_X$, but I can't quite see it.
If there is some other arrow $u_2:Prightarrow X$ such that $u_0circ u_2=1_P$ then I agree we would have a contradiction, since we then have $$p_0=p_0circ1_P=p_0circ u_0circ u_2=x_0circ u_2$$ and $u_1$ is unique with this property, but is the existence of $u_2$ necessary if $u_0circ u_1neq1_P$? Could the identity not hold and there be no such $u_2$?
(There are of course corresponding identities for $p_1$ and $x_1$, but they lead me to the same question.)
category-theory products
asked Jul 14 at 18:16
Alec Rhea
703515
asked Jul 14 at 18:16
Alec Rhea
703515
asked Jul 14 at 18:16
Alec Rhea
703515
asked Jul 14 at 18:16
Alec Rhea
703515
703515
1
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24
add a comment |Â
1
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24
1
1
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Since $P, p_0,p_1$ is a product diagram, any pair of arrows $Cto A, Cto B$ leads through it by a unique $Cto P$.
Now, for the pair $(x_0,x_1)$ this unique arrow is $u_0:Xto P$, and for the pair $(p_0,p_1)$, it is certainly $1_P$. But, taking the compositions, also $u_0circ u_1:Pto P$ leads $(p_0,p_1)$ through itself.
This means, by uniqueness, $u_0circ u_1=1_P$.
answered Jul 14 at 19:15
Berci
56.4k23570
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
add a comment |Â
up vote
3
down vote
This is a way to express the uniqueness property connected with products.
If $P$ serves as product of $A$ and $B$ with projections $p_0:Pto A$ and $p_1:Pto B$ then pair $(p_0,p_1)$ is a so-called monosource which means that on base of the equalities: $$p_0circ f=p_0circ gtext and p_1circ f=p_1circ g$$ you are allowed to conclude that: $$f=g$$
In your case we have: $$p_0circ u_0circ u_1=p_0circmathsfid_Ptext and p_1circ u_0circ u_1=p_1circmathsfid_P$$ so you are allowed to conclude that: $$u_0circ u_1=mathsfid_P$$
Similarly it can be found that: $$u_1circ u_0=mathsfid_X$$
This together proves that $u_0$ and $u_1$ are isomorphisms and are inverses of eachother.
answered Jul 14 at 19:01
drhab
86.7k541118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $P, p_0,p_1$ is a product diagram, any pair of arrows $Cto A, Cto B$ leads through it by a unique $Cto P$.
Now, for the pair $(x_0,x_1)$ this unique arrow is $u_0:Xto P$, and for the pair $(p_0,p_1)$, it is certainly $1_P$. But, taking the compositions, also $u_0circ u_1:Pto P$ leads $(p_0,p_1)$ through itself.
This means, by uniqueness, $u_0circ u_1=1_P$.
answered Jul 14 at 19:15
Berci
56.4k23570
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
add a comment |Â
up vote
2
down vote
accepted
Since $P, p_0,p_1$ is a product diagram, any pair of arrows $Cto A, Cto B$ leads through it by a unique $Cto P$.
Now, for the pair $(x_0,x_1)$ this unique arrow is $u_0:Xto P$, and for the pair $(p_0,p_1)$, it is certainly $1_P$. But, taking the compositions, also $u_0circ u_1:Pto P$ leads $(p_0,p_1)$ through itself.
This means, by uniqueness, $u_0circ u_1=1_P$.
answered Jul 14 at 19:15
Berci
56.4k23570
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $P, p_0,p_1$ is a product diagram, any pair of arrows $Cto A, Cto B$ leads through it by a unique $Cto P$.
Now, for the pair $(x_0,x_1)$ this unique arrow is $u_0:Xto P$, and for the pair $(p_0,p_1)$, it is certainly $1_P$. But, taking the compositions, also $u_0circ u_1:Pto P$ leads $(p_0,p_1)$ through itself.
This means, by uniqueness, $u_0circ u_1=1_P$.
answered Jul 14 at 19:15
Berci
56.4k23570
Since $P, p_0,p_1$ is a product diagram, any pair of arrows $Cto A, Cto B$ leads through it by a unique $Cto P$.
Now, for the pair $(x_0,x_1)$ this unique arrow is $u_0:Xto P$, and for the pair $(p_0,p_1)$, it is certainly $1_P$. But, taking the compositions, also $u_0circ u_1:Pto P$ leads $(p_0,p_1)$ through itself.
This means, by uniqueness, $u_0circ u_1=1_P$.
answered Jul 14 at 19:15
Berci
56.4k23570
answered Jul 14 at 19:15
Berci
56.4k23570
answered Jul 14 at 19:15
Berci
56.4k23570
answered Jul 14 at 19:15
Berci
56.4k23570
56.4k23570
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
add a comment |Â
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
1
1
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
Excellent, I hadn't thought of pulling a product diagram down over itself using the identity. Much appreciated!
– Alec Rhea
Jul 14 at 22:17
add a comment |Â
up vote
3
down vote
This is a way to express the uniqueness property connected with products.
If $P$ serves as product of $A$ and $B$ with projections $p_0:Pto A$ and $p_1:Pto B$ then pair $(p_0,p_1)$ is a so-called monosource which means that on base of the equalities: $$p_0circ f=p_0circ gtext and p_1circ f=p_1circ g$$ you are allowed to conclude that: $$f=g$$
In your case we have: $$p_0circ u_0circ u_1=p_0circmathsfid_Ptext and p_1circ u_0circ u_1=p_1circmathsfid_P$$ so you are allowed to conclude that: $$u_0circ u_1=mathsfid_P$$
Similarly it can be found that: $$u_1circ u_0=mathsfid_X$$
This together proves that $u_0$ and $u_1$ are isomorphisms and are inverses of eachother.
answered Jul 14 at 19:01
drhab
86.7k541118
add a comment |Â
up vote
3
down vote
This is a way to express the uniqueness property connected with products.
If $P$ serves as product of $A$ and $B$ with projections $p_0:Pto A$ and $p_1:Pto B$ then pair $(p_0,p_1)$ is a so-called monosource which means that on base of the equalities: $$p_0circ f=p_0circ gtext and p_1circ f=p_1circ g$$ you are allowed to conclude that: $$f=g$$
In your case we have: $$p_0circ u_0circ u_1=p_0circmathsfid_Ptext and p_1circ u_0circ u_1=p_1circmathsfid_P$$ so you are allowed to conclude that: $$u_0circ u_1=mathsfid_P$$
Similarly it can be found that: $$u_1circ u_0=mathsfid_X$$
This together proves that $u_0$ and $u_1$ are isomorphisms and are inverses of eachother.
answered Jul 14 at 19:01
drhab
86.7k541118
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is a way to express the uniqueness property connected with products.
If $P$ serves as product of $A$ and $B$ with projections $p_0:Pto A$ and $p_1:Pto B$ then pair $(p_0,p_1)$ is a so-called monosource which means that on base of the equalities: $$p_0circ f=p_0circ gtext and p_1circ f=p_1circ g$$ you are allowed to conclude that: $$f=g$$
In your case we have: $$p_0circ u_0circ u_1=p_0circmathsfid_Ptext and p_1circ u_0circ u_1=p_1circmathsfid_P$$ so you are allowed to conclude that: $$u_0circ u_1=mathsfid_P$$
Similarly it can be found that: $$u_1circ u_0=mathsfid_X$$
This together proves that $u_0$ and $u_1$ are isomorphisms and are inverses of eachother.
answered Jul 14 at 19:01
drhab
86.7k541118
This is a way to express the uniqueness property connected with products.
If $P$ serves as product of $A$ and $B$ with projections $p_0:Pto A$ and $p_1:Pto B$ then pair $(p_0,p_1)$ is a so-called monosource which means that on base of the equalities: $$p_0circ f=p_0circ gtext and p_1circ f=p_1circ g$$ you are allowed to conclude that: $$f=g$$
In your case we have: $$p_0circ u_0circ u_1=p_0circmathsfid_Ptext and p_1circ u_0circ u_1=p_1circmathsfid_P$$ so you are allowed to conclude that: $$u_0circ u_1=mathsfid_P$$
Similarly it can be found that: $$u_1circ u_0=mathsfid_X$$
This together proves that $u_0$ and $u_1$ are isomorphisms and are inverses of eachother.
answered Jul 14 at 19:01
drhab
86.7k541118
answered Jul 14 at 19:01
drhab
86.7k541118
answered Jul 14 at 19:01
drhab
86.7k541118
answered Jul 14 at 19:01
drhab
86.7k541118
86.7k541118
add a comment |Â
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%2f2851830%2funiqueness-of-categorical-products%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
1
Don't forget there is a corresponding pair of identities involving $x_1$ and $p_1$. (Incidentally, you may need to change your labeling of the maps.)
– Fan Zheng
Jul 14 at 18:24