When is the product of two injective functions, also injective? I'm currently, by curiosity, investigating when the product of two injective functions, is also injective. My condition is that this is true if and only if there $\nexists x_1, x_2 \in X$ such that $f(x_1) = g(x_2)$ and $g(x_1) = f(x_2)$. Here's my reasoning, I hope someone can confirm / disconfirm it, or leave some response.
Proof. Let's suppose we got two functions: $f: X \mapsto Y $, and $g: X \mapsto Y $, where the function $h(x)$ is defined as the product between the two:
$$h(x) = f(x)g(x)$$
The definition of an injective function, states that for some $x_1,x_2 \in X$ where $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$. Which then also holds for $g(x)$.
Let's use this fact for our newly defined function:
$$\left\{\begin{matrix}
h(x_1) = f(x_1)g(x_1)\\ 
h(x_2) = f(x_2)g(x_2)
\end{matrix}\right.$$
If $h(x)$ is injective, then: $x_1 \neq x_2 \Rightarrow h(x_1) \neq h(x_2)$. Because $f(x)$ and $g(x)$ are injective functions, the statement is true only and only if $f(x_1) \neq g(x_2)$ and $g(x_1) \neq f(x_2)$. $\ \ \ \ \ \ \ \ \ \square$
Can this be interpreted graphically, where the functions can't have any common y - values if I haven't misunderstood myself? Is there any other way to interpret the given result?
Thanks for your help!
 A: Your statement that $h$ is injective if and only if $\nexists x_1, x_1$ such that $f(x_1) = g(x_2)$ and $g(x_1) = f(x_2)$ is incorrect. Consider $f(x) = x$ and $g(x) = x + 2$.
Then $h(x) = x^2 + 2x$. Note that $h(0) = h(-2)$. So $h$ is not injective.
Now suppose we have $x_1, x_2$ such that $f(x_1) = g(x_2)$ and $f(x_2) = g(x_1)$. Then $x_1 = x_2 + 2$ and $x_2 = x_1 + 2$. Then $x_2 = x_2 + 4$. Contradiction.
A: Nope, it isn't injective. Counterexample:
Let be $f: \{0, 1\} \to \{-1, 1\}$ such that $f(0) = 1$ and $f(1) = -1$
$f^2(x)$ is the constant $1$, which is not injective for multiple possible inputs.
A: Multiplying functions is actually typically how functions fail to be injective.
Consider a simple example: $x \mapsto x$ is injective, but $x \mapsto x \cdot x = x^2$ is not. In fact, the latter is 2-to-1 at every point except $x = 0$.
More generally, if a function "looks like" $x \mapsto (x - c) \cdot (x - c) \cdots (x-c) = (x - c)^n$ near a point $c$, then in a sense that can be made precise, the function is locally $n$-to-1.
