Prove the following statement by contradiction:
Let $f:A \to B$. Then $f^{−1}:Range(f) \to A$ implies $f$ is one-to-one.
I can deduce why the statement must be true but i am having trouble using a proof by contradiction.
My proof so far assumes that $f$ is not one to one and $f^{−1}:Range(f) \to A$,
Which implies that the inverse function in not a function, which is a contradiction.
FSOC assume that $f:A \to B$ is not 1 to 1 and $f^{-1}:Ran(f) \to A$
$\therefore \exists y \in B | y=f(a_1 )=f(a_2 )$,where $a_1 \neq a_2$
$\therefore \exists a_1$ and $\exists a_2 \in Dom(f)$ where $a_1 \neq a_2 | f(a_1 )=f(a_2 )=y_1$
$(\therefore f^{-1} (Ran(f))$ is not a function because it does not pass the vertical line test.#)
Is this proof acceptable?
Yes, it is acceptable. Note that from your second point, it follows that $\exists y \in Range(f)$ such that $ f^{-1}(y) = a_{1} = a_{2}$, which contradicts the fact that $f^{-1}$ is a function.