## First Isomorphism Theorem for semigroups

Given a semigroup $S$ and a congruence $\sigma$ on $S$, show that the mapping $\pi_\sigma : S \rightarrow S/\sigma$ given by $s \mapsto [s]_\sigma$ is a surjective semigroup homomorphism.

Let $\varphi : S \rightarrow T$ be a homomorphism of semigroups. Define a relation $\mathsf{ker}\ \varphi$ by $(x,y) \in \mathsf{ker}\ \varphi$ if and only if $\varphi(x) = \varphi(y)$. Show that $\mathsf{ker}\ \varphi$ is a congruence on $S$. Show further that $S/\mathsf{ker}\ \varphi \cong \mathsf{im}\ \varphi$.

Recall that $S/\sigma$ is a semigroup under the operator $[x]_\sigma \star [y]_\sigma = [xy]_\sigma$, which is well-defined (i.e. does not depend on the representatives chosen) since $\sigma$ is a congruence. Now $\pi_\sigma(st) = [st]_\sigma = [s]_\sigma[t]_\sigma$ $= \pi_\sigma(s)\pi_\sigma(t)$, so that $\pi_\sigma$ is a semigroup homomorphism. Moreover, we have $[s]_\sigma = \pi_\sigma(s)$ for all $s \in S$, so that $\pi_\sigma$ is surjective.

Now let $\varphi$ and $\mathsf{ker}\ \varphi$ be as described above. Certainly $\mathsf{ker}\ \varphi$ is an equivalence. Now if $(x,y) \in \mathsf{ker}\ \varphi$ and $(z,w) \in \mathsf{ker}\ \varphi$, then $\varphi(x) = \varphi(y)$ and $\varphi(z) = \varphi(w)$, so that $\varphi(xz) = \varphi(x) \varphi(z) = \varphi(y) \varphi(w) = \varphi(yw)$. Hence $(xz,yw) \in \mathsf{ker}\ \varphi$, and so $\mathsf{ker}\ \varphi$ is a congruence on $S$.

Now define $\psi \subseteq S/\mathsf{ker}\ \varphi \times T$ by $\psi = \{ ([x]_\mathsf{ker}\ \varphi, \varphi(x)) \ |\ x \in S \}$. Certianly $\psi$ is total, and we claim it is well defined. Indeed, if we have $x,y \in S$ such that $(x,y) \in \mathsf{ker}\ \varphi$, then $\varphi(x) = \varphi(y)$. So in fact $\psi : S/\mathsf{ker}\ \varphi \rightarrow T$ is a function. Since $\psi([x][y]) = \psi([xy])$ $= \varphi(xy) = \varphi(x)\varphi(y)$ $= \psi([x])\psi([y])$, $\psi$ is a semigroup homomorphism. If $\psi([x]) = \psi([y])$, then we have $\varphi(x) = \varphi(y)$, so that $[x] = [y]$. Hence $psi$ is injective. Finally, if $t \in \mathsf{im}\ \varphi$, with $t = \varphi(s)$, then $t = \psi([s])$. So $\psi$ is onto the subset $\mathsf{im}\ \varphi \subseteq T$, and we have $S/\mathsf{ker}\ \varphi \cong \mathsf{im}\ \varphi$.