Can an object in a category have multiple distinct identity morphisms?












0












$begingroup$


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







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  • $begingroup$
    What is $id_1circ id_2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 29 at 5:05
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $id_1circ id_2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 29 at 5:05














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






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oldest

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3












$begingroup$

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.






share|cite|improve this answer









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  • $begingroup$
    :) nice. Thank you.
    $endgroup$
    – user56834
    Jan 29 at 5:14












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    :) nice. Thank you.
    $endgroup$
    – user56834
    Jan 29 at 5:14
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    :) nice. Thank you.
    $endgroup$
    – user56834
    Jan 29 at 5:14














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 at 5:10









Ivo TerekIvo Terek

46.6k954146




46.6k954146












  • $begingroup$
    :) 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




$begingroup$
:) nice. Thank you.
$endgroup$
– user56834
Jan 29 at 5:14


















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