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.