(Sorry for the formatting, I don't know how to include diagrams into stackexchange, so I just inserted the photos of the diagram which I made using LaTeX)
Recently I have been studying Paolo Aluffi's book, Algebra: Chapter 0.
Most of the constructions in the book regarding category theory felt like a breeze, until I met this thing, $\mathsf{C}^{\alpha, \beta}$.
In exercise 5.12, I was asked to define the notions of a fibered coproduct on $\mathsf{C}=\mathsf{Set}$, by considering the inital objects in $\mathsf{C}^{\alpha, \beta}$. Here was what I did.
Suppose $A, B, C \in \mathsf{Set}$, $\alpha, \beta$ are morphisms $\alpha: C \rightarrow A$, $\beta : C \rightarrow B$.
First define an relation ${\sim}$ on $A \bigsqcup B$ to be the "finest equivalence relation" such that $i_A\alpha(c) {\sim} i_B\beta(c)$ for any $c \in C$, $i_A,i_B$ are isomorphism from $A,B$ to their "isomorphic copies" in $A \bigsqcup B$ respectively.
Claim: The commutative diagram 
is an initial object in $\mathsf{C}^{\alpha, \beta}$, where $\pi$ is the canonical projection from $A \bigsqcup B$ onto $(A \bigsqcup B)/{\sim}$.
Proof:
It suffices to show that for any object in $\mathsf{C}^{\alpha, \beta}$,
,
there exists an unique commutative diagram 
Define $\sigma:(A \bigsqcup B)/{\sim} \rightarrow Z$ by $\sigma([s]_{{\sim}})=f_Ai_A^{-1}(s)$ if $s \in im~ i_A$ and $f_Bi_B^{-1}(s)$ if $s \in im~ i_B$.
I defined another relation $\thickapprox$ on $A \bigsqcup B$ to be the "finest equivalence relation" such that $s' \thickapprox s''$ iff $f_Ai_A^{-1}(s')$ or $f_Bi_B^{-1}(s')$ equals $f_Ai_A^{-1}(s'')$ or $f_Bi_B^{-1}(s'')$ (whichever pair is defined).
Note that for any $c \in C$, $i_A\alpha(c) \thickapprox i_B\beta(c)$ since $f_Ai_A^{-1}(i_A\alpha(c))=f_A\alpha(c)=f_B\beta(c)=f_Bi_B^{-1}(i_B\beta(c))$.
So ${\sim}$ must be "at least as fine" as $\thickapprox$ by construction.
Here I will have to prove that $\sigma$ is well-defined as a function.
To show that $\sigma$ is a well-defined function, I need to show that $[s']_{\sim}=[s'']_{\sim} \implies \sigma([s']_{\sim})=\sigma([s'']_{\sim})$.
$~~~~~~~~~[s']_{\sim}=[s'']_{\sim}$
$\implies s' {\sim} s''$
$\implies s' \thickapprox s''$ (since ${\sim}$ is "at least as fine" as $\thickapprox$)
$\implies f_Ai_A^{-1}(s')$ or $f_Bi_B^{-1}(s')$ equals $f_Ai_A^{-1}(s'')$ or $f_Bi_B^{-1}(s'')$ (whichever pair is defined)
$\implies \sigma(s') = \sigma (s'')$
So $\sigma$ is a function.
The commutativity of the diagram and uniqueness of $\sigma$ follow with a little bit more work.
I felt quite comfortable with this proof for a while, but then I started to wonder whether the "finest equivalence relation" in my definition of ${\sim}$ and $\thickapprox$ should even exist. Just as if the set $\{x \in \mathbb{R} | x > 3 \}$ doesn't have a least element.
Is there a more natural, less involved way to define the fibered coproduct, which would evade the current problems with my method?