Show coequalizers always exist in the collection $K$ of algebras of similar type with all homomorphisms between them

Question

Show coequalizers always exist in the collection of algebras of similar type with all homomorphisms between them

Definition: A coequalizer of two homomorphisms $h,k : A \to B$ in $K$ consists of another algebra $C$ and homomorphism $s : B \to C$ such that $sh = sk$ and given another algebra $D$ with homorphism $t : B \to D$ with $th = tk$ then there is a unique mapping $r : C \to D$ such that $rs = t$

If I understand correctly I need to prove this unique $r$ exists.

EDIT 2:

Let $A, B $ be similar algebras in $K$ such that $h,k : A \to B$ are homomorphisms

So I need to construct the pair $\langle C, s \rangle$ such that $s : B \to C$ such the following holds:

i) $sh = sk$

ii) for any $D$ in $K$ with homomorphism $t$ such that $t : B \to D$ and $th=tk$ then $\exists! rs = t$ where $r: C \to D$

Answer 1

Let $h,k : A \to B$ be homorphisms where $A,B \in K$

define $C = B/\theta$ where theta is the equivalence relation defined by $\forall a \in A, h(a) = k(a)$

we get that $sh = sk$ under this as $s(h(a)) = [h(a)] = [k(a)] = s(k(a))$

Now let $D$ in $K$ be such that $t : B \to D$ where $th =tk$ then define

$r : B/\theta \to D$ where $r([b]) = t(b)$ then we have $r(s(b)) = r[b] = t(b)$ so we have $rs = t$