Why $f(x)=x$ for any $x\in f(X)$?

Question

The set $S(X,X)$ of all mappings of a set $X$ to itself with the composition of mappings in the role of multiplication, where $|X|>1$. Why is not it a group?
Let $X$ be a nonempty set. Then the idempotents of the semigroup $S(X,X)$ of all mappings of $X$ to itself are precisely the mappings $f: X \to X$ satisfying the condition $f(x)=x$ for every $x \in f(X)$. An element $x$ of a semigoup is callled an idemotent if $xx=x$. My question is this: Here why $f(x)=x$ for any $x\in f(X)$?

Thanks a Lot!

I will be very frank: given the mathematical level of the questions you have asked here on the site before, I would have assumed you could easily answer these questions yourself. — Zev Chonoles, Sep 22 '13 at 6:45
@ZevChonoles: I'm a beginner of abstract-algebra and I want to be sure of something which I'm not sure. This is the purpose that I post question which seems that a little stupid. — Paul, Sep 22 '13 at 6:53

Zev Chonoles · Accepted Answer · 2013-09-22 06:28:06Z

1

Hints:

answered Sep 22 '13 at 6:28

Zev Chonoles

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$\begingroup$ 1, Why $f$ has not inverse? 2, I only have $f(f(x))=f(x)$. $\endgroup$ – Paul Sep 22 '13 at 6:30
$\begingroup$ Regarding 2: Saying that $$f(f(x))=f(x)$$ for all $x\in X$ is the same thing as saying that $$f(y)=y$$ for all $y\in f(X)$. $\endgroup$ – Zev Chonoles Sep 22 '13 at 6:32
$\begingroup$ Regarding 1: There will be some functions $f:X\to X$ that do not have inverses, and others that do. Do you understand what it means for a function to be injective? Do you see how can make a function $f:X\to X$ that is not injective if $X$ has more than one element? $\endgroup$ – Zev Chonoles Sep 22 '13 at 6:33

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