An Idempotent Continuous Function Pinned at Both Ends

Detailed Solution

Because $f(f(x))$ is defined for every $x\in[0,1]$ and the domain of $f$ is $[0,1]$, we necessarily have \[ f(x)\in[0,1]\qquad\text{for every }x\in[0,1]. \] Thus, \[ f([0,1])\subseteq[0,1]. \] Also, $f$ is continuous and \[ f(0)=0,\qquad f(1)=1. \] By the Intermediate Value Theorem, for every $y\in[0,1]$ there exists some $x\in[0,1]$ such that \[ f(x)=y. \] Therefore, \[ [0,1]\subseteq f([0,1]). \] Hence, \[ f([0,1])=[0,1]. \] Now let $y\in[0,1]$. Since $y$ belongs to the range of $f$, there exists $x\in[0,1]$ such that \[ f(x)=y. \] Using the given condition, \[ f(f(x))=f(x). \] Therefore, \[ f(y)=f(f(x))=f(x)=y. \] Since $y$ was arbitrary, \[ \boxed{f(y)=y\quad\text{for every }y\in[0,1]}. \] Thus, \[ \boxed{f(x)=x\quad\text{for all }x\in[0,1]}. \] Therefore, the correct answer is \[ \boxed{\mathrm{A}}. \] Why the other options are incorrect: - (B) is false because the conditions actually force $f$ to be the identity. The identity is strictly increasing, but the statement says that $f$ need not be the identity. - (C) is impossible because $f(0)=0$ and $f(1)=1$, so $f$ cannot be constant. - (D) is false because we have proved that every continuous function satisfying the given conditions must be the identity. Hence the unique correct answer is \[ \boxed{\mathrm{A}}. \]