Need help on function proof: If $\beta\alpha$ is one-to-one and $\alpha$ is onto, then $\beta$ is one-to-one. I just need help collecting my thoughts and maybe some hints on how to get started/continue.
What I know:
I know that $\beta\alpha$ being one-to-one implies that $\alpha$ is also one-to-one (already proved this in a previous exercise). So, if $\alpha$ is one-to-one and onto (as given), then $\alpha$ is a bijection and therefore $|A|=|B|$ and $\alpha^{-1}$ exists.
My attempt at a proof:
$\textbf{Proposition.}$ If $\beta\alpha$ is one-to-one and $\alpha$ is onto, then $\beta$ is one-to-one.
$\textit{Proof.}$ Let $\beta\alpha$ be one-to-one and $\alpha$ be onto. Also let $a_1\neq a_2$ where $a_i\in A$. Because $\beta\alpha$ is one-to-one, $\alpha$ must also be one-to-one. Due to this fact, $\alpha(a_1)\neq \alpha(a_2)$. Also note that $\beta\alpha(a_1)\neq \beta\alpha(a_2)$ due to $\beta\alpha$ being one-to-one. By definition, $\beta\alpha(a_1)\neq \beta\alpha(a_2)$ is equivalent to $\beta[\alpha(a_1)] \neq \beta[\alpha(a_2)]$. Letting $\alpha(a_i)=b_i$, $\beta[\alpha(a_1)] \neq \beta[\alpha(a_2)]$ is equivalent to $\beta[b_1] \neq \beta[b_2]$. Because $b_1 \neq b_2$ implies that $\beta[b_1] \neq \beta[b_2]$, then $\beta$ is one-to-one.
 A: The proof strikes me as quite unclear and uses some irrelevant facts - although it possibly has the right idea hidden in the second to last sentence. I think it would largely be fixed by starting out with a more clear definition of your givens and your claim. You might start, for instance, as:

Let $\alpha$ be a function $A\rightarrow B$ and $\beta$ be a function $B\rightarrow C$ such that $\alpha$ is onto and $\beta\alpha$ is one-to-one. We will show that $\beta$ is also one-to-one.

Where we clearly state our goals, including the domains and codomains of the functions (since we will need to discuss these later). After such an introduction, you should proceed to expand the definition of "one-to-one" - so, you might continue as follows*

Suppose that $b_1$ and $b_2$ are distinct members of $B$. We claim that $\beta(b_1)\neq \beta(b_2)$.

This is essentially fixed: there is not really any other way to start this proof. Your proof starts by fixing elements of $A$ - but that has nothing to do with whether $\beta$ is one-to-one, so is not a good way to start.
You have this claim in your proof that showing $\beta(b_1)\neq \beta(b_2)$ is the same as showing that $\beta\alpha(a_1)=\beta\alpha(a_2)$ for some particular values of $a_1$ and $a_2$ - and this is the key idea - but you need to phrase it better, and show how such values can be derived from the values $b_1$ and $b_2$. For instance, you could continue:

Since $\alpha$ is onto, we may choose values $a_1$ and $a_2$ in $A$ such that
$$\alpha(a_1)=b_1$$
$$\alpha(a_2)=b_2$$

And then, from this stage, you can observe that $\beta\alpha(a_1)=\beta(b_1)$ and $\beta\alpha(a_2)=\beta(b_2)$ - and then I'll leave it to you to finish this proof using the fact that $\beta\alpha$ is one-to-one.
