Fibered coproducts in $\mathsf{Set}$ Following my not-entirely-successful attempt to define fibered products in $\mathsf{Set}$, I will attempt to define the fibered coproducts:
Let $A,B,C$ be sets, and let $\alpha\colon C\to A$ and $\beta\colon C\to B$. I want to find the fibered coproduct of $\alpha$ and $\beta$.
Let $Z$ be a set and let $f\colon A\to Z$ and $g\colon B\to Z$ with $f\alpha=g\beta$. By the universal property of disjoint unions, we can combine $f$ and $g$ to form a function $h\colon A\sqcup B\to Z$. Now because $f\alpha=g\beta$, $h(\iota_A(\alpha(c)))=h(\iota_B(\beta(c)))$, inducing a symmetric relation on $A\sqcup B$ whose transitive closure, $\sim$, is respected by any such $h$. Let $\pi\colon A\sqcup B\to (A\sqcup B)/\sim$ be the projection. Then $\phi_A:=\pi \iota_A$ and $\phi_B:=\pi\iota_B$ form the fibered coproduct:
By the universal property of the quotient, there is a unique $h'\colon (A\sqcup B)/\sim$ such that $h=h'\pi$ which occurs when $f=h\iota_A=h'\pi\iota_A=h'\phi_A$ and $g=h'\phi_B$.

Is this correct? Is there a better way?
 A: The argument I gave essentially contains a sketch of a proof that coequalizers in $\mathsf{Set}$ are as described in Malice Vidrine's comment. All that remains is to show that a category with (binary) coproducts and coequalizers has (binary) fibered coproducts. The general finite and infinite cases look to be about the same.
Let $\alpha\colon C\to A$ and $\beta\colon C\to B$ as before. Let $e\colon A\amalg B\to D$ be the coequalizer of $i_A \alpha$ and $i_B \beta$. By the definition of coproduct, $hi_A=f$ and $hi_B=g$. By the definition of coequalizer, $(ei_A)\alpha=e(i_A\alpha)=e(i_B\beta)=(ei_B)\beta$. I wish to show that in fact $ei_A\colon A\to D$ and $ei_B\colon B\to D$ form the fibered coproduct of $\alpha$ and $\beta$.
Let $f\colon A\to Z$, $g\colon B\to Z$, and $f\alpha=g\beta$. Then by the definition of coproduct, there is a unique $h\colon A\amalg B\to Z$ such that $f=hi_A$ and $g=hi_B$. Then by the definition of coequalizer, there is a unique $h'\colon D\to Z$ such that $h'e=h$, so $h'$ is the unique morphism from $D$ to $Z$ such that $f=h'(ei_A)$ and $g=h'(ei_B)$.
Does this look right?
