# Repeated Composition of Injection is Injection

## Theorem

Let $S$ be a set.

Let $f: S \to S$ be an injection.

- $f^0, f^1, f^2, \ldots, f^n, \ldots$

be defined as:

- $\forall n \in \N: \map {f^n} x = \begin {cases} x & : n = 0 \\ \map f x & : n = 1 \\ \map f {\map {f^{n - 1} } x} & : n > 1 \end{cases}$

Then for all $n \in \N$, $f^n$ is an injection.

## Proof

Proof by induction:

For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition:

- $f^n$ is an injection.

$\map P 0$ is true, as this is the case Identity Mapping is Injection.

$\map P 1$ is true, as this is the assertion that $f$ is an injection.

### Basis for the Induction

$\map P 2$ is the assertion that $f^2$ is an injection.

\(\ds \map {f^2} x\) | \(=\) | \(\ds \map f {\map f x}\) | Definition of $f^n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\paren {f \circ f} } x\) | Definition of Composition of Mappings |

But $f \circ f$ is an injection by Composite of Injections is Injection.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $f^k$ is an injection.

Then we need to show:

- $f^{k + 1}$ is an injection.

### Induction Step

This is our induction step:

\(\ds \map {f^{k + 1} } x\) | \(=\) | \(\ds \map f {\map {f^k} x}\) | Definition of $f^n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\paren {f \circ f^k} } x\) |

By the induction hypothesis $f^k$ is an injection.

By Composite of Injections is Injection it follows that $f^{k + 1}$ is an injection.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore for all $n \in \N$:

- $f^n$ is an injection.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $3$