# Set theory - Inclusion map vs identity map

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...

• So, $X\ne X{}{}$? – Lord Shark the Unknown Aug 19 '18 at 11:41
• You mean like the inclusion map of one set into another set that properly contains it? – rschwieb Aug 19 '18 at 11:42
• You denote two distinct sets both by $X$. That is confusing. – drhab Aug 19 '18 at 11:55
• drhab it's the way it's written in the book :) – Paul Aug 19 '18 at 12:02
• What book is this from, Paul? – Carl Mummert Aug 19 '18 at 12:06

## 2 Answers

The inclusion map $A\to B$ 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 $A\to B$ 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.

• Thanks for your answer. Can you please give me an example of the different codomains? – Paul Aug 19 '18 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 '18 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 '18 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 '18 at 12:20
• Okay thanks for your help...still confused though – Paul Aug 19 '18 at 12:29

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

The maps $\Bbb R\to \Bbb R:x\mapsto |x|$ and $\Bbb R\to \Bbb R_+:x\mapsto |x|$ are two different maps.

If $A\subset B$,

• the inclusion map is defined as $A\to B : x\mapsto x$,

• the identity map is defined as $A\to A : x\mapsto x$.

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

• 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 '18 at 12:43