This is exercise 2.4.viii of Riehl's "Category Theory in Context":
Prove that for any $F:\rm{C} \to \rm{Set}$, the canonical forgetful functor $\prod:\int F\to \rm{C}$ has the following property: for any morphism $f:c\to d$ in $\rm{C}$ and any object $(c,x)$ in the fiber over $c$, there is a unique lift of the morphism $f$ to a morphism in $\int F$ with domain $(c,x)$ that projects along $\prod$ to $f$.
I know that the map $(c,x) \rightarrow (d, Ff(x))$ would project to $f$ through $\prod$, but I can't quite see why it is unique. I'm not sure why there couldn't be a different morphism $g : c \to d$ in $\rm{C}$ such that $Fg(x)=Ff(x) = y$.
The only thing I can think of is that even if such a map exists, the corresponding map $(c,x) \rightarrow (d, y)$ is the same as the one created by $f$ in $\int F$. Is this correct? And if it is, does this imply that there is at most one morphism between any two objects in $\int F$?
Thanks in advance!
This is a matter of chasing the definitions.
A morphism in $\int F$ with domain $(c,x)$ has the form $g : (c,x) \to (c',x')$, where $g : c \to c'$ is a morphism in $C$, $x' \in F(c')$ and $F(g)(x)=x'$.
By definition, $\Pi(g : (c,x) \to (c',x'))$ is $g : c \to c'$. It is required that this is equal to $f : c \to d$. This means $c'=d$ and $g = f$. Also, $x'=F(g)(x)=F(f)(x)$. So everything is uniquely determined.