Why is $\operatorname{colim} F \cong \pi_0\left (\int F\right )$? Given a small functor $F:\mathsf{C \to Set}$, I need to prove that $\operatorname{colim} F$ is isomorphic/in bijection with the connected components of the category of elements $\int F$.  It's not the least bit clear to me how to make this connection, a number of approaches have led me to dead ends and I just can't see where to go.  I would just like a gentle nudge since I'm getting extremely frustrated with myself for not being able to solve this.
Observation #1
Recognize that $\int F$ is the pullback of the following equalizer diagram in $\mathsf{CAT}$:
$$
\require{AMScd}
\begin{CD}
\int F @>>> \mathsf{Set}_* \\
@V{\Pi}VV   @VUVV \\
\mathsf{C} @>>F> \mathsf{Set}
\end{CD}
$$
where the top arrow is inclusion.  I'm not sure how to translate this into a statement about colimits, but I do know that we can consider the pushout diagram afforded to us by the Yoneda embedding:
$$
\require{AMScd}
\begin{CD}
\mathsf{CAT}(\mathsf{Set}, \cdot ) @>{U^*}>> \mathsf{CAT}(\mathsf{Set}_*, \cdot ) \\
@V{F^*}VV   @VVV \\
\mathsf{CAT}(\mathsf{C}, \cdot ) @>>{\Pi^*}> \mathsf{CAT}\left (\int F, \cdot \right )
\end{CD}
$$
where morphisms in $\mathsf{CAT}(\mathsf{A, B})$ are functors $\mathsf{A \to B}$.  Can something be done here where we can argue a representation of Cone$(F, \cdot)$?  My understanding of what to do with this information is eluding me.
Observation #2
I suppose anything else I might know of what to do here would involve trying to construct an isomorphism directly between the colimit in question and connected components of $\int F$, yet these connected components are confusing me as well.  Is there some canonical way of picking an object $c \in \mathsf{C}$ and $x \in Fc$ such that the equivalence class $[(c,x)]$ will naturally yield some unique value elements of $\operatorname{colim} F$?
I am utterly lost here.  On a personal note, I'm frankly embarrassed that I can't see the crucial observations needed to prove this statement.  It feels like something I should be able to see since it appears to be stated as a straightforward exercise.  Maybe I'm simply not understanding some general approach to computing colimits.  Sorry for venting my impostor syndrome; I know it's outside the scope of this site and not protected content.
 A: I think I finally saw the key observation needed in this.  For any diagram $F: \mathsf{C\to Set}$ and cone $\lambda: F \Longrightarrow X$ for $X \in \mathsf{Set}$ we have individual morphisms $\lambda_c: Fc \to X$ indexed by the objects of $\mathsf{C}$ and the compatibility condition that states for any $f \in \mathsf{C}(c,d)$ we have that $\lambda_c = \lambda_d\circ Ff$.
The key observation comes in seeing each leg of the cone $\lambda_c: Fc \to X$ as really a slice of the mapping $\lambda: \int F \to X$.  In other words, an element $x \in Fc$ can really be thought of as the pair $(c,x) \in \int F$ and $\lambda_c(x) = \lambda(c,x)$.  Most importantly the compatibility condition shows that any two elements $(c,x)$ and $(d,y)$ in $\int F$ get mapped to the same element of $X$ as long as there's a morphism $f \in \mathsf{C}(c,d)$ where $Ff(x) = y$.  By extension, any two elements of $\int F$ joined by a finite sequence of such morphisms gets mapped to the same element of $X$.  Thus we can conclude $\lambda:\int F \to X$ is constant on the path components of $\int F$ and therefore descends to a mapping $\tilde{\lambda}: \pi_0\left (\int F\right ) \to X$.  This mapping is unique since a different mapping would not be compatible with the cone $\lambda$ specified above; i.e. this is the only such mapping that $\lambda$ can descend to.
Furthermore, this induces a cone $\eta:F \Longrightarrow \pi_0\left (\int F\right )$ which assigns $x \in Fc$ its path component $[(c,x)]$.  We see that any cone $\lambda:F\Longrightarrow X$ factors uniquely as $\lambda = \tilde{\lambda}\circ \eta$.
A: The proof you give in your answer is the one I would give. Just for fun, here's a much more roundabout proof that I like. We need a bunch of adjunctions:

*

*$\pi_0\colon\mathrm{Cat}\rightarrow\mathrm{Set}$ is right adjoint to the incousion $\mathrm{Set}\subset\mathrm{Cat}$;

*$\int \colon\mathrm{Fun}(C,\mathrm{Set})\rightarrow\mathrm{Cat}/C$ is right adjoint to the functor $L$ given by $L(p:D\rightarrow C)(c) = \pi_0 (p/c)$;

*The forgetful functor $\pi_!\colon \mathrm{Cat}/C\rightarrow\mathrm{Cat}$ is left adjoint to $D\mapsto (C\times D\rightarrow C)$.

In fact the first adjunction is a special case of the second. We also need a couple facts:

*

*The composite $L\circ \int$ is naturally equivalent to the identity on $\mathrm{Fun}(C,Set)$;

*If $c_X\colon C\rightarrow \mathrm{Set}$ is constant on a set $X$, then $\int c_X = (C\times X\rightarrow C)$.

The hard work is hiding in verifying the stated relations between $L$ and $\int$. Now the diagram
$$
\require{AMScd}
\begin{CD}
\mathrm{Cat}/C @>{\pi_!}>> \mathrm{Cat} \\
@V{L}VV   @VV{\pi_0}V \\
\mathrm{Fun}(C,\mathrm{Set}) @>>\operatorname{colim}> \mathsf{Set}
\end{CD}
$$
commutes, for these are all left adjoints, and the diagram of right adjoints commutes by the preceding comments. So for $F\colon C\rightarrow\mathrm{Set}$, we can compute
$$\operatorname{colim} F = \operatorname{colim} L \int F = \pi_0 \pi_! \int F = \pi_0 \int F.$$
If you run this argument not for $\pi_!\colon \mathrm{Cat}/C\rightarrow\mathrm{Cat}$, but for $p_!\colon \mathrm{Cat}/C\rightarrow\mathrm{Cat}/E$ with some given $p\colon C\rightarrow E$, you recover the colimit formula for left Kan extensions.
