Set theory - Inclusion map vs identity map

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I'm trying to wrap my head around inclusion maps and identity maps...



if X is a subset of Y, the function f defined by f(x)=x for each x in X is called the inclusion map of X into Y.



If X=1,2,3 and Y=1,2,3,4,5. Then f=(1,1),(2,2),(3,3)?



The inclusion map of X into X is called identity map on X. (In the language of relations, the identity map on X is the same as the relation of equality in X)



So in my example the inclusion map = identity map. Can someone give me an example of where this is not true?



Thanks in advance...







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  • 5




    So, $Xne X$?
    – Lord Shark the Unknown
    Aug 19 at 11:41










  • You mean like the inclusion map of one set into another set that properly contains it?
    – rschwieb
    Aug 19 at 11:42











  • You denote two distinct sets both by $X$. That is confusing.
    – drhab
    Aug 19 at 11:55










  • drhab it's the way it's written in the book :)
    – Paul
    Aug 19 at 12:02










  • What book is this from, Paul?
    – Carl Mummert
    Aug 19 at 12:06














up vote
0
down vote

favorite
1












I'm trying to wrap my head around inclusion maps and identity maps...



if X is a subset of Y, the function f defined by f(x)=x for each x in X is called the inclusion map of X into Y.



If X=1,2,3 and Y=1,2,3,4,5. Then f=(1,1),(2,2),(3,3)?



The inclusion map of X into X is called identity map on X. (In the language of relations, the identity map on X is the same as the relation of equality in X)



So in my example the inclusion map = identity map. Can someone give me an example of where this is not true?



Thanks in advance...







share|cite|improve this question


















  • 5




    So, $Xne X$?
    – Lord Shark the Unknown
    Aug 19 at 11:41










  • You mean like the inclusion map of one set into another set that properly contains it?
    – rschwieb
    Aug 19 at 11:42











  • You denote two distinct sets both by $X$. That is confusing.
    – drhab
    Aug 19 at 11:55










  • drhab it's the way it's written in the book :)
    – Paul
    Aug 19 at 12:02










  • What book is this from, Paul?
    – Carl Mummert
    Aug 19 at 12:06












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I'm trying to wrap my head around inclusion maps and identity maps...



if X is a subset of Y, the function f defined by f(x)=x for each x in X is called the inclusion map of X into Y.



If X=1,2,3 and Y=1,2,3,4,5. Then f=(1,1),(2,2),(3,3)?



The inclusion map of X into X is called identity map on X. (In the language of relations, the identity map on X is the same as the relation of equality in X)



So in my example the inclusion map = identity map. Can someone give me an example of where this is not true?



Thanks in advance...







share|cite|improve this question














I'm trying to wrap my head around inclusion maps and identity maps...



if X is a subset of Y, the function f defined by f(x)=x for each x in X is called the inclusion map of X into Y.



If X=1,2,3 and Y=1,2,3,4,5. Then f=(1,1),(2,2),(3,3)?



The inclusion map of X into X is called identity map on X. (In the language of relations, the identity map on X is the same as the relation of equality in X)



So in my example the inclusion map = identity map. Can someone give me an example of where this is not true?



Thanks in advance...









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 19 at 12:32

























asked Aug 19 at 11:40









Paul

155




155







  • 5




    So, $Xne X$?
    – Lord Shark the Unknown
    Aug 19 at 11:41










  • You mean like the inclusion map of one set into another set that properly contains it?
    – rschwieb
    Aug 19 at 11:42











  • You denote two distinct sets both by $X$. That is confusing.
    – drhab
    Aug 19 at 11:55










  • drhab it's the way it's written in the book :)
    – Paul
    Aug 19 at 12:02










  • What book is this from, Paul?
    – Carl Mummert
    Aug 19 at 12:06












  • 5




    So, $Xne X$?
    – Lord Shark the Unknown
    Aug 19 at 11:41










  • You mean like the inclusion map of one set into another set that properly contains it?
    – rschwieb
    Aug 19 at 11:42











  • You denote two distinct sets both by $X$. That is confusing.
    – drhab
    Aug 19 at 11:55










  • drhab it's the way it's written in the book :)
    – Paul
    Aug 19 at 12:02










  • What book is this from, Paul?
    – Carl Mummert
    Aug 19 at 12:06







5




5




So, $Xne X$?
– Lord Shark the Unknown
Aug 19 at 11:41




So, $Xne X$?
– Lord Shark the Unknown
Aug 19 at 11:41












You mean like the inclusion map of one set into another set that properly contains it?
– rschwieb
Aug 19 at 11:42





You mean like the inclusion map of one set into another set that properly contains it?
– rschwieb
Aug 19 at 11:42













You denote two distinct sets both by $X$. That is confusing.
– drhab
Aug 19 at 11:55




You denote two distinct sets both by $X$. That is confusing.
– drhab
Aug 19 at 11:55












drhab it's the way it's written in the book :)
– Paul
Aug 19 at 12:02




drhab it's the way it's written in the book :)
– Paul
Aug 19 at 12:02












What book is this from, Paul?
– Carl Mummert
Aug 19 at 12:06




What book is this from, Paul?
– Carl Mummert
Aug 19 at 12:06










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










The inclusion map $Ato B$ is always the same set of pairs as the identity map on $A$.



So if you're in a context where maps are simply sets of pairs (and, in particular, a map has a range but does not have a unique codomain), then they are the same.



On the other hand, at least informally (and in quite a number of formal contexts too) it is usual and convenient to speak of a map as "knowing" what its domain and codomain are. In that case the inclusion map $Ato B$ differs from the identity map, because (assuming $B$ is a proper superset of $A$) their codomains are different -- even though both have the same domain and they have the same value at every point in the domain.



This perspective is needed, for example, if we want to ask whether the map is surjective or not. The identity map is a surjection; a nontrivial inclusion map is not.






share|cite|improve this answer






















  • Thanks for your answer. Can you please give me an example of the different codomains?
    – Paul
    Aug 19 at 12:06










  • @Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
    – Henning Makholm
    Aug 19 at 12:09










  • Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
    – Paul
    Aug 19 at 12:12










  • @Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
    – Henning Makholm
    Aug 19 at 12:20











  • Okay thanks for your help...still confused though
    – Paul
    Aug 19 at 12:29

















up vote
0
down vote













A map is not defined only by a set of arrows $xmapsto f(x)$, it is defined by a domain and a codomain.



The maps $Bbb Rto Bbb R:xmapsto |x|$ and $Bbb Rto Bbb R_+:xmapsto |x|$ are two different maps.



If $Asubset B$,



  • the inclusion map is defined as $Ato B : xmapsto x$,


  • the identity map is defined as $Ato A : xmapsto x$.


The output is the same, but the codomain isn't.






share|cite|improve this answer




















  • This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
    – Carl Mummert
    Aug 19 at 12:43











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










The inclusion map $Ato B$ is always the same set of pairs as the identity map on $A$.



So if you're in a context where maps are simply sets of pairs (and, in particular, a map has a range but does not have a unique codomain), then they are the same.



On the other hand, at least informally (and in quite a number of formal contexts too) it is usual and convenient to speak of a map as "knowing" what its domain and codomain are. In that case the inclusion map $Ato B$ differs from the identity map, because (assuming $B$ is a proper superset of $A$) their codomains are different -- even though both have the same domain and they have the same value at every point in the domain.



This perspective is needed, for example, if we want to ask whether the map is surjective or not. The identity map is a surjection; a nontrivial inclusion map is not.






share|cite|improve this answer






















  • Thanks for your answer. Can you please give me an example of the different codomains?
    – Paul
    Aug 19 at 12:06










  • @Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
    – Henning Makholm
    Aug 19 at 12:09










  • Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
    – Paul
    Aug 19 at 12:12










  • @Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
    – Henning Makholm
    Aug 19 at 12:20











  • Okay thanks for your help...still confused though
    – Paul
    Aug 19 at 12:29














up vote
2
down vote



accepted










The inclusion map $Ato B$ is always the same set of pairs as the identity map on $A$.



So if you're in a context where maps are simply sets of pairs (and, in particular, a map has a range but does not have a unique codomain), then they are the same.



On the other hand, at least informally (and in quite a number of formal contexts too) it is usual and convenient to speak of a map as "knowing" what its domain and codomain are. In that case the inclusion map $Ato B$ differs from the identity map, because (assuming $B$ is a proper superset of $A$) their codomains are different -- even though both have the same domain and they have the same value at every point in the domain.



This perspective is needed, for example, if we want to ask whether the map is surjective or not. The identity map is a surjection; a nontrivial inclusion map is not.






share|cite|improve this answer






















  • Thanks for your answer. Can you please give me an example of the different codomains?
    – Paul
    Aug 19 at 12:06










  • @Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
    – Henning Makholm
    Aug 19 at 12:09










  • Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
    – Paul
    Aug 19 at 12:12










  • @Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
    – Henning Makholm
    Aug 19 at 12:20











  • Okay thanks for your help...still confused though
    – Paul
    Aug 19 at 12:29












up vote
2
down vote



accepted







up vote
2
down vote



accepted






The inclusion map $Ato B$ is always the same set of pairs as the identity map on $A$.



So if you're in a context where maps are simply sets of pairs (and, in particular, a map has a range but does not have a unique codomain), then they are the same.



On the other hand, at least informally (and in quite a number of formal contexts too) it is usual and convenient to speak of a map as "knowing" what its domain and codomain are. In that case the inclusion map $Ato B$ differs from the identity map, because (assuming $B$ is a proper superset of $A$) their codomains are different -- even though both have the same domain and they have the same value at every point in the domain.



This perspective is needed, for example, if we want to ask whether the map is surjective or not. The identity map is a surjection; a nontrivial inclusion map is not.






share|cite|improve this answer














The inclusion map $Ato B$ is always the same set of pairs as the identity map on $A$.



So if you're in a context where maps are simply sets of pairs (and, in particular, a map has a range but does not have a unique codomain), then they are the same.



On the other hand, at least informally (and in quite a number of formal contexts too) it is usual and convenient to speak of a map as "knowing" what its domain and codomain are. In that case the inclusion map $Ato B$ differs from the identity map, because (assuming $B$ is a proper superset of $A$) their codomains are different -- even though both have the same domain and they have the same value at every point in the domain.



This perspective is needed, for example, if we want to ask whether the map is surjective or not. The identity map is a surjection; a nontrivial inclusion map is not.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 19 at 12:03

























answered Aug 19 at 11:48









Henning Makholm

229k16294525




229k16294525











  • Thanks for your answer. Can you please give me an example of the different codomains?
    – Paul
    Aug 19 at 12:06










  • @Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
    – Henning Makholm
    Aug 19 at 12:09










  • Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
    – Paul
    Aug 19 at 12:12










  • @Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
    – Henning Makholm
    Aug 19 at 12:20











  • Okay thanks for your help...still confused though
    – Paul
    Aug 19 at 12:29
















  • Thanks for your answer. Can you please give me an example of the different codomains?
    – Paul
    Aug 19 at 12:06










  • @Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
    – Henning Makholm
    Aug 19 at 12:09










  • Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
    – Paul
    Aug 19 at 12:12










  • @Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
    – Henning Makholm
    Aug 19 at 12:20











  • Okay thanks for your help...still confused though
    – Paul
    Aug 19 at 12:29















Thanks for your answer. Can you please give me an example of the different codomains?
– Paul
Aug 19 at 12:06




Thanks for your answer. Can you please give me an example of the different codomains?
– Paul
Aug 19 at 12:06












@Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
– Henning Makholm
Aug 19 at 12:09




@Paul: $A=1,2,3$ and $B=1,2,3,4,5$ would seem to be an excellent example.
– Henning Makholm
Aug 19 at 12:09












Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
– Paul
Aug 19 at 12:12




Hmm I'm still confused as to how the codomains are different. I guess I need to do some more study, I'm new to this...
– Paul
Aug 19 at 12:12












@Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
– Henning Makholm
Aug 19 at 12:20





@Paul: They're defined to be different. The identity map by definition has domain $A$ and codomain $A$; the injection map by definition has domain $A$ and codomain $B$.
– Henning Makholm
Aug 19 at 12:20













Okay thanks for your help...still confused though
– Paul
Aug 19 at 12:29




Okay thanks for your help...still confused though
– Paul
Aug 19 at 12:29










up vote
0
down vote













A map is not defined only by a set of arrows $xmapsto f(x)$, it is defined by a domain and a codomain.



The maps $Bbb Rto Bbb R:xmapsto |x|$ and $Bbb Rto Bbb R_+:xmapsto |x|$ are two different maps.



If $Asubset B$,



  • the inclusion map is defined as $Ato B : xmapsto x$,


  • the identity map is defined as $Ato A : xmapsto x$.


The output is the same, but the codomain isn't.






share|cite|improve this answer




















  • This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
    – Carl Mummert
    Aug 19 at 12:43















up vote
0
down vote













A map is not defined only by a set of arrows $xmapsto f(x)$, it is defined by a domain and a codomain.



The maps $Bbb Rto Bbb R:xmapsto |x|$ and $Bbb Rto Bbb R_+:xmapsto |x|$ are two different maps.



If $Asubset B$,



  • the inclusion map is defined as $Ato B : xmapsto x$,


  • the identity map is defined as $Ato A : xmapsto x$.


The output is the same, but the codomain isn't.






share|cite|improve this answer




















  • This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
    – Carl Mummert
    Aug 19 at 12:43













up vote
0
down vote










up vote
0
down vote









A map is not defined only by a set of arrows $xmapsto f(x)$, it is defined by a domain and a codomain.



The maps $Bbb Rto Bbb R:xmapsto |x|$ and $Bbb Rto Bbb R_+:xmapsto |x|$ are two different maps.



If $Asubset B$,



  • the inclusion map is defined as $Ato B : xmapsto x$,


  • the identity map is defined as $Ato A : xmapsto x$.


The output is the same, but the codomain isn't.






share|cite|improve this answer












A map is not defined only by a set of arrows $xmapsto f(x)$, it is defined by a domain and a codomain.



The maps $Bbb Rto Bbb R:xmapsto |x|$ and $Bbb Rto Bbb R_+:xmapsto |x|$ are two different maps.



If $Asubset B$,



  • the inclusion map is defined as $Ato B : xmapsto x$,


  • the identity map is defined as $Ato A : xmapsto x$.


The output is the same, but the codomain isn't.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 19 at 12:28









Arnaud Mortier

19.6k22159




19.6k22159











  • This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
    – Carl Mummert
    Aug 19 at 12:43

















  • This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
    – Carl Mummert
    Aug 19 at 12:43
















This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
– Carl Mummert
Aug 19 at 12:43





This is a common convention, but Halmos uses a slightly different convention, in which those two are the same map. Instead of saying something like "$f$ is surjective", he always says "$f$ maps $X$ onto $Y$", viewing surjectivity as a property of $f$ and $Y$ together rather than as a property of $f$ alone. Similarly, in Halmos' conventions, a map from $X$ to $Y$ is also a map from $X$ to $Z$ whenever $Y subseteq Z$.
– Carl Mummert
Aug 19 at 12:43













 

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