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!
Best Answer
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.