# $f:X \to X$, a bijection, and $A \subseteq X$, Assume $f(A) \subseteq A$, then $f|_A$ is a bijection [closed]

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
• 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) Apr 2 at 3:06
• 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? Apr 2 at 3:15
• Domain will be A and the codomain will be f(A), which is a subset of A here... Apr 2 at 6:15
• If the codomain is $f(A)$, then it's trivially onto (regardless of whether it's a subset of $A$ or not). Apr 2 at 7:17
• 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.) Apr 2 at 16:15

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.