# Injective function between a monoid and a function from same monoid to monoid

Suppose we have $$M$$ a monoid.

Define $$E(M) = \{\alpha : M\rightarrow M : \alpha(xy)=\alpha(x) \cdot y \}$$

If $$a \in M$$, define $$\alpha_{a}: M \rightarrow M$$ by \begin{align*} \alpha_{a}(x)=ax \quad \forall x \in M \end{align*}

Question is to prove that the function $$\theta: M \rightarrow E(M)$$ defined by $$\theta(a)=\alpha_{a} \quad \forall a \in M$$ is injective.

My attempt:

Suppose for some $$a, b \in M$$ \begin{align*} \theta(a)=\theta(b) \Rightarrow \alpha_{a}=\alpha_{b} & \Rightarrow \forall x \in M, ax=bx \end{align*}

I am now stuck on this part. In a group, I would take the inverse of $$x$$, however it is not necessary the case that every element has an inverse. I guess the key question that I am asking is whether in every row/column of a Cayley table of a monoid, does each element occur exactly once?

Suppose that, for all $$x \in M$$, $$ax = bx$$. Taking $$x = 1$$, the identity of the monoid, you get $$a = b$$. Thus $$\theta$$ is injective.