I have just started learning about groups and rings and I'm stuck on one exercise. I don't understand what $S^u$ really is and don't know where to start. So if anybody could help me with it, it would be amazing.
The exercise is the following:
Let $S$ be a semigroup, and let $u$ be an element such that $u \notin S$. On the disjoint union $S \sqcup\{u\}$ we define a binary operation $$\cdot \colon (S \sqcup\{u\}) \times(S \sqcup\{u\}) \rightarrow(S \sqcup\{u\})$$ given by: $$ \left \{ \begin{array}[cc] ss \cdot t=s t & \text{if } s, t \in S , \\ s \cdot u=u \cdot s=s & \text{if } s \in S, \\ u \cdot u=u & \end{array} \right. $$ The resulting semigroup $S^{u}$ is a monoid. We define the unitization $\hat{S}$ of $S$ as $$\hat{S}= \left \{ \begin{array}[cc] SS & \text{if } S \text{ is a monoid} \\ S^{u} & \text{otherwise} \end{array} \right. $$
Prove that if $M$ is a monoid and $\alpha: S \rightarrow M$ is a homomorphism, there exists a unique homomorphism of monoids $\hat{\alpha}: \hat{S} \rightarrow M$ such that $\hat{\alpha}(x)=\alpha(x)$ if $x \in S$.
If $S$ is a monoid, then $\hat S = S$ and $\hat \alpha = \alpha$. Otherwise, $\hat S = S^u$. Let $1$ be the identity of $M$. If $\hat \alpha$ exists, it is necessarily defined by $$ (*) \quad \hat \alpha(s) = \begin{cases} \alpha(s) &\text{if $s \in S\quad$ (by definition)} \\ 1 &\text{if $s = u\quad$ (by definition of a homomorphism of monoids)} \end{cases} $$ This proves unicity. It remains to prove that $(*)$ defines a homomorphism of monoids. Let $s, t \in S^u$. Then $$ \hat\alpha(s)\hat\alpha(t) = \begin{cases} \alpha(s)\alpha(t) = \alpha(st) = \hat\alpha(st) &\text{if $s,t \in S$}\\ \alpha(s)1 = \alpha(s) = \alpha(su) = \hat\alpha(st) &\text{if $s \in S$ and $t=u$}\\ 1\alpha(t) = \alpha(t) = \alpha(ut) = \hat\alpha(st) &\text{if $s = u$ and $t \in S$}\\ 1 = \hat\alpha(st) & \text{if $s = t = u$}\end{cases} $$ and thus $\hat \alpha(s)\hat\alpha(t) = \hat\alpha(st)$ in all cases. Moreover $\hat\alpha(u) = 1$ by definition.