Assume that alpha: S-T, and beta: T-U, and gamma: T-U. Prove the following statement

If beta of alpha is invertible, then beta is onto and alpha is one-to-one

Printable View

- Sep 7th 2008, 11:14 AMmandy123prove
Assume that alpha: S-T, and beta: T-U, and gamma: T-U. Prove the following statement

If beta of alpha is invertible, then beta is onto and alpha is one-to-one - Sep 7th 2008, 01:45 PMJhevon
i might as well give a full solution. the proof for this is all over the internet and is in most textbooks that deal with functions indepth. it might as well be on here...or ami just re-inventing the wheel? ...or maybe bored.

**Proof:**

We proceed using a direct proof. Assume $\displaystyle \beta \circ \alpha$ is invertible. Then $\displaystyle \beta \circ \alpha$ is one-to-one and onto (there is a theorem that tells us this, it should be in your text).

We show $\displaystyle \beta$ is onto: Let $\displaystyle z \in U$. Since $\displaystyle \beta \circ \alpha $ is onto, there exists $\displaystyle x \in S$ such that $\displaystyle (\beta \circ \alpha)(x) = z$. But that means $\displaystyle \beta (\alpha (x)) = z$ with $\displaystyle \alpha (x) \in T$. Thus $\displaystyle \beta$ is onto.

We show $\displaystyle \alpha$ is one-to-one: Assume $\displaystyle x_1, x_2 \in S$ with $\displaystyle \alpha (x_1) = \alpha (x_2)$. Then $\displaystyle \beta (\alpha (x_1)) = \beta (\alpha (x_2))$, or in other words, $\displaystyle (\beta \circ \alpha)(x_1) = (\beta \circ \alpha)(x_2)$. But $\displaystyle \beta \circ \alpha$ is one-to-one, so $\displaystyle (\beta \circ \alpha)(x_1) = (\beta \circ \alpha)(x_2) \implies x_1 = x_2$. Hence, $\displaystyle \alpha$ is one-to-one

**QED**