Endomorphism monoid

Remarks

The composite of two homomorphisms is a homomorphism and $\id_A$ is a homomorphism, so $E$ is closed under $\circ$ and contains $\id_A$; composition is associative.

Example: let $\mathbf{A} = (\Z, +, 0)$. Each endomorphism $h$ satisfies $h(x) = x\,h(1)$ for all $x \in \Z$, by induction on $x \ge 0$ and $h(-x) = -h(x)$. So $h \mapsto h(1)$ is an isomorphism from $\End(\mathbf{A})$ to the monoid $(\Z, \cdot, 1)$, since $(g \circ h)(1) = g(1)\,h(1)$.

Non-example: the map $x \mapsto x + 1$ lies in the monoid of all maps $\Z \to \Z$ but not in $\End(\Z, +, 0)$, because it sends $0$ to $1$.

The invertible elements of $\End(\mathbf{A})$ are the automorphisms of $\mathbf{A}$.