Mixing
Can an object in a category have multiple distinct identity morphisms?
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The definition of a category requires that there exists a morphism for each object $X$, $id_X$ such that for all morphisms $f$, $fcirc id_X=id_Xcirc f = f$.
But it doesn't say that this morphism is unique.
Is an object in a category allowed to have multiple, distinct morphisms $id_1,id_2,...$ that all satisfy that for all morphisms $f$, $fcirc id_i=id_icirc f = f$?
category-theory
asked Jan 29 at 5:02
user56834user56834
3,37921253
$endgroup$
add a comment |
$begingroup$
The definition of a category requires that there exists a morphism for each object $X$, $id_X$ such that for all morphisms $f$, $fcirc id_X=id_Xcirc f = f$.
But it doesn't say that this morphism is unique.
Is an object in a category allowed to have multiple, distinct morphisms $id_1,id_2,...$ that all satisfy that for all morphisms $f$, $fcirc id_i=id_icirc f = f$?
category-theory
asked Jan 29 at 5:02
user56834user56834
3,37921253
$endgroup$
$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05
add a comment |
$begingroup$
The definition of a category requires that there exists a morphism for each object $X$, $id_X$ such that for all morphisms $f$, $fcirc id_X=id_Xcirc f = f$.
But it doesn't say that this morphism is unique.
Is an object in a category allowed to have multiple, distinct morphisms $id_1,id_2,...$ that all satisfy that for all morphisms $f$, $fcirc id_i=id_icirc f = f$?
category-theory
asked Jan 29 at 5:02
user56834user56834
3,37921253
$endgroup$
The definition of a category requires that there exists a morphism for each object $X$, $id_X$ such that for all morphisms $f$, $fcirc id_X=id_Xcirc f = f$.
But it doesn't say that this morphism is unique.
Is an object in a category allowed to have multiple, distinct morphisms $id_1,id_2,...$ that all satisfy that for all morphisms $f$, $fcirc id_i=id_icirc f = f$?
category-theory
category-theory
asked Jan 29 at 5:02
user56834user56834
3,37921253
asked Jan 29 at 5:02
user56834user56834
3,37921253
asked Jan 29 at 5:02
user56834user56834
3,37921253
asked Jan 29 at 5:02
user56834user56834
3,37921253
asked Jan 29 at 5:02
user56834user56834
3,37921253
3,37921253
$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05
add a comment |
$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05
$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05
$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05
add a comment |
1 Answer
1
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$begingroup$
You have that $${rm Id}_1 stackrel{(1)}{=} {rm Id}_1 circ {rm Id}_2 stackrel{(2)}{=} {rm Id}_2,$$where in $(1)$ we use that ${rm Id}_2$ is a "right-identity" and on (2) that ${rm Id}_1$ is a "left-identity". Just like for groups.
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
$endgroup$
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
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
$begingroup$
You have that $${rm Id}_1 stackrel{(1)}{=} {rm Id}_1 circ {rm Id}_2 stackrel{(2)}{=} {rm Id}_2,$$where in $(1)$ we use that ${rm Id}_2$ is a "right-identity" and on (2) that ${rm Id}_1$ is a "left-identity". Just like for groups.
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
$endgroup$
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
add a comment |
$begingroup$
You have that $${rm Id}_1 stackrel{(1)}{=} {rm Id}_1 circ {rm Id}_2 stackrel{(2)}{=} {rm Id}_2,$$where in $(1)$ we use that ${rm Id}_2$ is a "right-identity" and on (2) that ${rm Id}_1$ is a "left-identity". Just like for groups.
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
$endgroup$
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
add a comment |
$begingroup$
You have that $${rm Id}_1 stackrel{(1)}{=} {rm Id}_1 circ {rm Id}_2 stackrel{(2)}{=} {rm Id}_2,$$where in $(1)$ we use that ${rm Id}_2$ is a "right-identity" and on (2) that ${rm Id}_1$ is a "left-identity". Just like for groups.
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
$endgroup$
You have that $${rm Id}_1 stackrel{(1)}{=} {rm Id}_1 circ {rm Id}_2 stackrel{(2)}{=} {rm Id}_2,$$where in $(1)$ we use that ${rm Id}_2$ is a "right-identity" and on (2) that ${rm Id}_1$ is a "left-identity". Just like for groups.
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
answered Jan 29 at 5:10
Ivo TerekIvo Terek
46.6k954146
46.6k954146
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
add a comment |
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14
add a comment |
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$begingroup$
What is $id_1circ id_2$?
$endgroup$
– Lord Shark the Unknown
Jan 29 at 5:05