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Let $f:X \to X$ be a bijection, where $A \subseteq X$. Now if we asssume that $f(A) \subseteq A$, then $f|_A$ is a bijection

It is easy to prove that $f$ is $1$ to $1$, but I am unable to prove that f is Onto.

I would appreciate any help.

Note: This is a wrong assumption, counterexample given by @davidlui in answer
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    $\begingroup$ Usually, $f|_A : A \to X$, which wouldn't be a bijection unless $A=X$. Even if you define $f|_A : A \to A$, it may not be. The closest you can get is it is a bijection onto its image (i.e. the same as being an injection) $\endgroup$ Apr 2 at 3:06
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    $\begingroup$ To be explicit, my point with the previous comment is this: how do you define $f|_A$? What is the domain? What is the codomain? $\endgroup$ Apr 2 at 3:15
  • $\begingroup$ Domain will be A and the codomain will be f(A), which is a subset of A here... $\endgroup$ Apr 2 at 6:15
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    $\begingroup$ If the codomain is $f(A)$, then it's trivially onto (regardless of whether it's a subset of $A$ or not). $\endgroup$ Apr 2 at 7:17
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    $\begingroup$ If $A$ is a finite set, though, then it is true. (Because in general, a one-to-one function from a finite set to itself is a bijection.) $\endgroup$ Apr 2 at 16:15

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This is not true. Let $X = \mathbb{Z}$ and $f(x) = x+1$. Let $A = \{n \in \mathbb{Z} : n \geq 0\}$. Then, $f$ is a bijection, $f(A) \subseteq A$, but $f|_A$ is not a bijection.

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