Show that $\sigma$ is a bijection if and only $u$ is a unit.

Question

The full exercise asks

If $M$ is a monoid and $u \in M$, let $\sigma: M \to M$ be defined by $\sigma(a) = ua$ for all $a \in M$. Show that $\sigma$ is a bijection if and only $u$ is a unit.

The backward part of this exercise was easy. But I have no idea how to prove the forward part of the statement, which requires proving that $u$ is a unit. (I just got back from a vacation, so I still hasn't gotten used to doing math again and I guess that's one reason why I'm struggling so much lol). I've been stuck on this for like 40 minutes and I have no idea what to do. Some help (preferably a hint) would be appreciated!

Here is my proof for the backward part:

Suppose that $u$ is a unit. We start by showing that $\sigma$ is injective. Suppose that $\sigma (a) = \sigma (b)$, or $ua = ub$. Since $u$ is a unit, we may multiply both sides by $u^{-1}$ to get $a=b$. Next, let $a \in M$, and let $b=u^{-1} a$. Then $\sigma(b) = u b = u u^{-1} a = a$. Thus $\sigma$ is surjective, and hence bijective.

Edit: All this time I reversed the forward and backward statements of every "if and only if statement." Sometimes I ask myself how I've gotten this far.

Answer 1

Here's the proof that if $\sigma$ is a bijection then $u$ is a unit.

First, $\sigma$ is surjective. So $1$ is in the image: there is some $x$ such that $\sigma (x) = ux = 1$. $x$ is our candidate for the inverse of $u$. We know it's a right inverse, but we also have to show that it's a left inverse.

Consider $xu$. We have to show that this is equal to $1$. If we apply $\sigma$ to it, we get $\sigma(xu)=uxu= u$. However, we also have that $\sigma (1) = u$. $\sigma$ is injective, so $xu = 1$.

Therefore the picked-out $x$ is an inverse on both sides, so it is $u^{-1}$ and $u$ is a unit.

Answer 2

If $u$ is a unit, you wish to show that $\sigma$ is a bijection. Let us first show that it is surjective. Indeed, let $x\in M$, since $u$ is a unit there is some $u^{-1}$ such that $uu^{-1}=1$. Thus, we have that $\sigma(u^{-1}x)=uu^{-1}x=x$, so we have that $\sigma$ is surjective. Now to show that $\sigma$ is injective. Say that we have that $\sigma(x)=\sigma(y)$, then we must have that $ux=uy$, and multiplying by $u^{-1}$ on the left of both sides we conclude that $x=y$, so $\sigma$ is injective completing the proof that if $u$ is a unit, then $\sigma$ is bijective.