Wikipedia says $i$ is an inclusion means $i: A \to B$ with $A \subset B$ means $i(x) = x$ for each $x\in A$.
But doesn't this mean $i(A) = A$, so this is actually identity?
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- 4$\begingroup$ To be the identity map it is necessary $A=B.$ $\endgroup$mfl– mfl2014-11-10 16:43:54 +00:00Commented Nov 10, 2014 at 16:43
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4 Answers
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6 No it isn't the identity since it isn't surjective. In the case when $B=A$ then this injection is the identity.
- $\begingroup$ But it is surjective with $A$. $\endgroup$Lemon– Lemon2014-11-10 16:45:37 +00:00Commented Nov 10, 2014 at 16:45
- $\begingroup$ What do you mean by surjective with $A$? $\endgroup$user63181– user631812014-11-10 16:46:23 +00:00Commented Nov 10, 2014 at 16:46
- $\begingroup$ I mean $i(A) = A$. $\endgroup$Lemon– Lemon2014-11-10 16:47:40 +00:00Commented Nov 10, 2014 at 16:47
- 4$\begingroup$ $i$ is surjective if $\forall b\in B$ there's $a\in A$ such that $i(a)=b$. Now take $b\in B-A$ to see that $i$ isn't surjective if $B\ne A$. $\endgroup$user63181– user631812014-11-10 16:50:41 +00:00Commented Nov 10, 2014 at 16:50
- $\begingroup$ Okay I understand that it isn't the identity on $B$, but it is identity on $A$ right? $\endgroup$Lemon– Lemon2014-11-10 16:52:05 +00:00Commented Nov 10, 2014 at 16:52
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8 Recall the definition: Two functions are equal if they have the same domain, codomain and if for each $x$ in the domain they take the same value...
You are missing the bolded part.
- 3$\begingroup$ I thought that two functions are equal if they are the same set of ordered pairs. $\endgroup$2014-11-10 17:07:59 +00:00Commented Nov 10, 2014 at 17:07
- 1$\begingroup$ @Asaf: While that definition has some merit, if you don't include the codomain as part of the definition, you can't speak of surjectivity as an intrinsic property of the function. (You can, of course, still speak of a function being "surjective over the set $B$".) $\endgroup$Ilmari Karonen– Ilmari Karonen2014-11-10 19:49:29 +00:00Commented Nov 10, 2014 at 19:49
- 1$\begingroup$ @Ilmari: Strange how set theorists talk about surjections all the time! $\endgroup$2014-11-10 20:02:17 +00:00Commented Nov 10, 2014 at 20:02
- $\begingroup$ @AsafKaragila Correct me if I am wrong, but I thought that the formal definition of a function is a SUBset of ordered pairs, the subset being in the product of domain and co-domain. $\endgroup$N. S.– N. S.2014-11-10 23:53:56 +00:00Commented Nov 10, 2014 at 23:53
- $\begingroup$ The two most common definitions are (1) a set of ordered pairs with some property, (2) an ordered triplet consisting of a set of ordered pairs, domain and codomain. In the first definition the domain can be extracted from the ordered pairs, but the codomain cannot, whereas in the second definition changing the codomain will change the function. Both definitions are useful, depending on what you are doing. $\endgroup$2014-11-11 03:25:49 +00:00Commented Nov 11, 2014 at 3:25
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Map i is not identity map,to be identity codomain and domain has to be same.
Also, we can say map i is identity on A.
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2 Two maps $f,g$ are the same if their domain is the same and $f(x)=g(x) \hspace{0.5cm}\forall \hspace{0.2cm}x \in D=D(f)=D(g)$. In your case the first condition is not satisfied.
- $\begingroup$ In the most common definition, the codomains are also required to be the same. In the original example, the functions have the same domain A, but different codomains (A vs B). $\endgroup$J-mster– J-mster2014-11-11 12:38:16 +00:00Commented Nov 11, 2014 at 12:38
- $\begingroup$ Thank you, I had not realize that in some context it may be appropiate to make this distinction. $\endgroup$somebody– somebody2014-11-16 11:47:35 +00:00Commented Nov 16, 2014 at 11:47