Problem with showing that the fibered product over a final object is the product Let $\mathcal C$ be a category where $X,Y,Z$ live and $Z$ is a final object. The fibered product of $X$ and $Y$ over $Z$ with given morphisms $\alpha: X\rightarrow Z$, $\beta: Y \rightarrow Z$
is the final object in a new category, $\mathcal D$, where objects are diagrams
$$ W \rightarrow X \xrightarrow{\alpha}Z = W \rightarrow Y \xrightarrow{\beta}Z $$
for $W \in \mathcal C$,
and morphisms in $\mathcal D$ are morphisms $f:W_1 \rightarrow W_2$
such that
$$ 
W_1 \xrightarrow{f} W_2 \rightarrow X \xrightarrow{\alpha}Z = W_2 \rightarrow X \xrightarrow{\beta} Z
$$
[At least, this is what I think the right definition is, since if I set the definition of morphisms in $\mathcal D$ to be morphisms $f: W_1 \rightarrow W_2$ satisfying
\begin{align*} W_1 \xrightarrow{f} W_2 \rightarrow X &= W_2 \rightarrow X \\
 W_1 \xrightarrow{f} W_2 \rightarrow Y &= W_2 \rightarrow Y
\end{align*}
then the fibered product would be the the product in general, not just for $Z$ the final object (wouldn't it?) ]
So, supposing that I have the right definition of a fibered product, supposing $Z$ is final, what I need to show is that for any $W \in \mathcal D$ i.e. $W \rightarrow Y$ and $W \rightarrow X$ with the commutativity condition, we have a unique morphism in $\mathcal D$
$$ W \rightarrow X \times Y$$
i.e.
$$ W \rightarrow X \times Y \rightarrow X \xrightarrow{\alpha} Z = X \times Y \rightarrow X \xrightarrow{\alpha} Z$$
Obviously, be the universal property of the product we have such a morphism $f: W \rightarrow X \times Y$, which satisfies an even stronger condition, namely:
\begin{align*}
W \xrightarrow{f} X \times Y \rightarrow X &= W \rightarrow X \\
W \xrightarrow{f} X \times Y \rightarrow Y &= W \rightarrow Y
\end{align*}
and $f$ is the unique morphism satisfying this condition.
But I'm at a loss as to showing that $f$ is the unique morphism which satisfies the (weaker) condition that
$$ W \xrightarrow{f} X \times Y \rightarrow X \xrightarrow{\alpha} Z = X \times Y \rightarrow X \xrightarrow{\beta} Z$$
Not only do I not see how the fact that $Z$ is a final object helps, but I think it hinders, since what this fact says is that if $g:W \rightarrow X \times Y$ is any morphism, then automatically
$$ W \xrightarrow{g} X \times Y \rightarrow X \xrightarrow{\alpha} Z = W \xrightarrow{f} X \times Y \rightarrow X \xrightarrow{\alpha} Z $$
since there is only one morphism from $W$ to $Z$! Whereas, to show that $f=g$, I would've liked to use the universal property of the product, and so I would've liked to show that
\begin{align*}
W \xrightarrow{g} X \times Y \rightarrow X &= W \rightarrow X \\
W \xrightarrow{g} X \times Y \rightarrow Y &= W \rightarrow Y
\end{align*}
but in general I don't see why this would be true of any morphism $g: W \rightarrow X \times Y$.
So... what gives? Am I messing up something silly? Is my definition of the fibered product wrong?
 A: Your definition of the fiber product is incorrect; it's the other thing, the one you said makes it the product in general (it doesn't). With the correct definition you use the fact that $Z$ is the final object so by definition every object has a unique morphism with target $Z$.
Anyway, I think you're making life a little too complicated for yourself. It's easiest to think instead about the universal property, or equivalently to think about the functor that $X \times_Z Y$ represents: by definition, a morphism $W \to X \times_Z Y$ is a pair of morphisms $W \to X, W \to Y$ such that the composites $W \to X \to Z, W \to Y \to Z$ are equal. That's it. If $Z$ is a final object then this second condition is automatic because, again, every object has a unique morphism with target $Z$. So the universal property is the same as the universal property of the product and you conclude by the Yoneda lemma.
If you haven't done it yet, compute the fiber product in $\text{Set}$. That will give you some sense of what it means (it is literally the fiberwise product, that's why it's called that, and if $Z$ is the final object there's only one fiber and it's the whole thing), and once you have that you can check that sense of meaning against any particular argument by specializing it to $\text{Set}$ and seeing what happens there.
