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Let $X$ be a finite set and let $f:X\rightarrow X$ be injective. Prove that $f$ is also surjective.

Here's my atempt:

Since $X$ is finite, there exists $g: I_n \rightarrow X$ bijective (Notation: $I_n = \{x\in N ; 1\leq x \leq n\}$). Since $f$ is injective, it follows that $f'=X \rightarrow f(X)$ is a bijective function. Noticing that $f(X)\subset X$, and that X is finite, it is also true that $f(X)$ is finite. By injectivity, we have that $h: I_n \rightarrow f(X)$ is also bijective. Let $\varphi$ be the following composition:

$\varphi = f'^{-1}\circ h \circ g^{-1} : X \rightarrow X$.

Since $f',g,h$ are all bijections, it follows that $\varphi$ is also bijective.

Now here come's my doubt: How do I guarantee/prove that $f$ and $\varphi$ are equal? Because if i prove that it's proven that $f$ is bijection.

3 Answers 3

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Maybe slightly easier:

Suppose $f$ is not surjective, meaning there is some $y \in X$ such that $f(y) \neq x$ for all $y \in X$. Thus $|f(X)| \leq |X| - 1$. On the other hand, we know that $f$ is injective, which means $|f(X)| = |X|$. This is a contradiction, and hence $f$ must be surjective.

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You want to prove that $|f(X)| = |X|$.

Suppose $|f(X)| < |X|$, then the pigeon hole principle shows that there is some $y \in f(X)$ such that $|f^{-1} (\{y\})| > 1$, which contradicts injectivity.

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    What happened? Why the unaccept? – 2017-01-23
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Why don't we proceed by induction? If we know that $f: X\to X$, $f$ injective, is a bijection for all $X$ of cardinality $n$, we can show the same for all $X$ of cardinality $n + 1$: let $g: Y \to Y$ be injective, where $|Y| = n+1$. Let $Y' \subset Y$ be any set of cardinality $n$. By induction, $g|_{Y'}$ is a bijection, so because $g$ is injective, there is only one place for the $y\not\in Y'$ to go!

Sanity check: what's the base case? Other sanity check: I left something out of this proof that is actually crucial; where did I omit a step? (Hint: what is image of $g|_{Y'}$?) How can you remedy this?