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I have been trying following question and was unable to solve.

Let $f: X \to X$ such that $f(f(x)) = x$ for all $x\in X. \space$ Then:

  1. Is $f$ 1-1?
  2. Is $f$ onto?

Clearly $f$ is 1-1 . But I am unable to deduce why $f$ must be onto or not.

Anton Vrdoljak
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  • Hint: for every $y\in X$, you're looking for an element $z\in X$ such that $f(\underline z)=y$. Does it help to use the assumption that $f(\underline{f(y)})=y$? – Greg Martin Aug 02 '20 at 15:39
  • Such a function is called an involution. As for surjectivity (onto-ness)... you just need to show that for each $x\in X$ there is some $y\in X$ such that $f(y)=x$ where such a $y$ might depend on $x$. Clearly, letting $y=f(x)$ works. – JMoravitz Aug 02 '20 at 15:40

3 Answers3

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If $f$ were not onto, since $f:X\to X$, $f(X)\subset X$ and this containment is proper. Then $f(f(X))\subset f(X)\subset X$, but this contradicts $f(f(x))=x$ for every $x\in X$.

Integrand
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If $z$ is an arbitrary element of $X$ then map it to get $f(z)$ which complies $f(f(z))=z$, so $f(z)$ is the preimage of $z$. Hence $f$ is surjective.

janmarqz
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$f$ is its own inverse, so $f$ is bijective and in particular surjective.

J. De Ro
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