Let $m , n\in \mathbb{Z_+}$. Let $X \neq \emptyset$. If $m \leq n$ find an injective map, $f: X^m \to X^n$.

Question

Let $m , n\in \mathbb{Z_+}$. Let $X \neq \emptyset$ If $m \leq n$ find an injective map, $f: X^m \to X^n$.

My Attempted Solution

If I define $f : X^m\to X^n$ such that $f(x_1 , ..x_m) = (x_1, ...x_m, \underbrace{a, ... ,a}_\text{$m-n$ times})$ where $a \in X$, then we can easily see that $f$ is injective.

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But I feel that my proposed solution to this problem is somewhat handwavy and non-rigorous. Is there a more rigorous way to define what I stated above, or is there a different cleaner way to construct an injective map given these conditions?

Answer 1

The idea is fine, and the only improvement in the presentation that I would suggest is to rearrange the proof to pick $a$ first: logically, you need $a$ before you can define $f$.

Fix $a\in X$, and for each $\langle x_1,\ldots,x_m\rangle\in X^m$ let

$$f(\langle x_1,\ldots,x_m\rangle)=\langle x_1,\ldots,x_m,\underbrace{a,\ldots,a}_{n-m}\rangle\;.$$

An alternative that avoids the underbrace:

Fix $a\in X$, and for $k=m+1,\ldots,n$ let $x_k=a$. Then define $f:X^m\to X^n$ by

$$f(\langle x_1,\ldots,x_m\rangle)=\langle x_1,\ldots,x_n\rangle$$

for each $\langle x_1,\ldots,x_m\rangle\in X^m$.