functions - $f:Xrightarrow X$ such that $f(f(x))=x$

Let $X$ be a metric space and $f:X\rightarrow X$ be such that $f(f(x))=x$, for all $x\in X$.

Then $f$

1. is one-one and onto;


2. is one-one but not onto;



3. is onto but not one-one;


4. need not be either.

From the given condition I have that $f^2=i$ where $i$ is the identity function. If $f$ itself is the identity function then the conditions are satisfied as well as $f$ is bijection. Is that the only such function or are there other possibilities ?

My guess is that it will be bijection i.e. option $1$ will be correct .

For see, if

$$f(x_1)=y \ \text{and}\ f(x_2)=y \ \text{then} \ f(y)=x_1\ \text{and}\ f(y)=x_2$$ will be possible iff $x_1=x_2$. So this is injective.

Now an injection from a set to itself is trivially surjective so it is bijective. Is my proof correct?