Composition and inverse mappings

Question

Let $A\stackrel{\alpha} \rightarrow B \stackrel{\beta}\rightarrow A$ satisfy $\beta \alpha = 1_A$. If either $\alpha$ is onto or $\beta$ is one to one, show that each of them is invertible and that each of them is the inverse of the other.

I have a few theorems that I am working with trying to prove this.

1. If $\alpha:A\rightarrow B$ has an inverse, the inverse mapping is unique.

2. $\alpha:A\rightarrow B$ and $\beta:B\rightarrow C$ are mappings

   a. $1_A:A\rightarrow A$ is invertible and $1_A^{-1}=1_A$

   b. if $\alpha$ is invertible, then $a^1$ is invertible and $(\alpha^{-1})^{-1}=\alpha$

   c. if $\alpha$ and $\beta$ are both invertible then $\beta\alpha$ is invertible

3. A mapping $\alpha: A\rightarrow B$ is invertible if and only if it is a bijection.

So my thinking is that if I can show that both $\alpha$ and $\beta$ are bijections then by 3 they are both invertible. Is there a way to do so from the information I have been given and will that be a sufficient proof?

Answer 1

Yes, the easiest way to prove this statement is to show that that $\alpha$ and $\beta$ both have an inverse function. As stated, they are actually each others inverse, so you must prove that $\beta\alpha = 1_A$ and $\alpha \beta = 1_B$ both hold. The first identitiy is one of the assumptions, so you only need to prove the second one.

Hint: Consider the functions $\alpha \beta \alpha$ respectively $\beta \alpha \beta$ and remember that the composition of functions is associative.

Answer 2

If we know that the composition $\beta \alpha$ of two functions $\alpha$ and $\beta$ is onto, then this implies that $\beta$ has to be onto. Similarly, if we know that $\beta \alpha$ is one to one, then $\alpha$ has to be one to one.

With these statements you should be able to prove what you want. As additional exercise, try to prove the statements I gave.