# Surjectivity implies injectivity

Let S be a finite set.Let F be a surjective function from S to S.

How do I prove that it is injective?

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 Have you tried counting elements yet? – Sebastian Sep 9 '11 at 10:32 Suppose $x \neq y \in S$ and that $f(x) =f(y)$. Let $|S|=n$. How many distinct elements can lie in the image of $f$? – mt_ Sep 9 '11 at 10:36

Let $S$ be a finite set, and $f : S \to S$ a function. Then the following are equivalent:

• $f$ is injective.
• $f$ is surjective.
• $f$ is bijective.

This is really just a counting argument. First, suppose $f$ is injective. If $S$ has $n$ elements, by our assumption, this means the image of $f$ has at least $n$ elements. But the image of $f$ is contained in $S$, so it has at most $n$ elements; so the image of $f$ contains exactly $n$ elements and is therefore the whole of $S$, i.e. $f$ is surjective.

Next, suppose $f$ is surjective. So, for each $y$ in $S$, there is an $x$ in $S$ such that $y = f(x)$; we choose one such $x$ for each $y$ and define a function $g : S \to S$ so that $g(y) = x$. By construction, $f(g(y)) = y$, so $g$ must be injective, and hence, must be surjective by the above argument. So $g$ is a bijection, and $f$ is a left inverse for $g$. But a left inverse for a bijection is also a right inverse, so this implies $f$ is a bijection, and a fortiori an injection.

Notice that the very first part of the argument fails when $S$ is not finite. For example, let us consider the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x + 1$. This function is certainly injective but is not surjective. Similarly, the function $g : \mathbb{N} \to \mathbb{N}$ defined by $f(0) = 0$ and $f(x + 1) = x$ is surjective, but not injective.

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 Why is the function g injective? – Mohan Sep 9 '11 at 11:06 @user774025: Because we send $y$ to its $x$ such that $f(x)=y$. Since $f$ is a function there can only be one element as $f(x)$. – Asaf Karagila Sep 9 '11 at 11:46