TITLE: Equivalence of universal properties
QUESTION [7 upvotes]: In "Sheaves in Geometry and Logic" Mac Lane, Moerdijk (page 292: section VI, chapter 4), say:
In a topos $\mathcal{E}$ (with terminal object $1$), take a family of objects $\{ F_{n} \}_{n \in \mathbb{N}}$ and the map
$\require{AMScd}$
\begin{CD}
\coprod_{n \in \mathbb{N}} F_{n} @>p>> \coprod_{n \in \mathbb{N}} 1\\
\end{CD}
induced by the unique maps
$\require{AMScd}$
\begin{CD}
F_{n} @>>> 1\\
\end{CD}
By the fact that, for any object $X$,
$$(\coprod_{n \in \mathbb{N}} 1) \times X \cong \coprod_{n \in \mathbb{N}} (1 \times X) \cong \coprod_{n \in \mathbb{N}} X$$
we can show that a map
$\require{AMScd}$
\begin{CD}
(\coprod_{n \in \mathbb{N}} 1) \times X @>g>> \coprod_{n \in \mathbb{N}} F_{n}\\
\end{CD}
such that $p \circ g = \pi_{1}$ (the projection on the component $\coprod_{n \in \mathbb{N}} 1$) it's just given by a family of maps
$\require{AMScd}$
\begin{CD}
X @>g_{n}>> \coprod_{n \in \mathbb{N}} F_{n}\\
\end{CD}
such that, by identity $p \circ g = \pi_{1}$, each $g_{n}$ factors through the component $F_{n}$.
The question is why the identity $p \circ g = \pi_{1}$ is equivalent to the fact that $g_{n}$ factors through $F_{n}$.
The only thing I discovered is that $\pi_{1}$ corresponds, via the isomorphism
$$(\coprod_{n \in \mathbb{N}} 1) \times X \cong \coprod_{n \in \mathbb{N}} X$$
to the map
$\require{AMScd}$
\begin{CD}
\coprod_{n \in \mathbb{N}} X @>>> \coprod_{n \in \mathbb{N}} 1\\
\end{CD}
induced by
$\require{AMScd}$
\begin{CD}
X @>>> 1 @>i_{n}>> \coprod_{n \in \mathbb{N}} 1\\
\end{CD}
in the $n$-th component.

REPLY [4 votes]: $\require{AMScd}$
The composition $X_i \to 1_i \to \coprod1$ equals the composition $X_i \to \coprod F_n \to \coprod 1,$ so it's enough to show that $F_i$ is the pullback
$\begin{CD}
F_i @>>> \coprod F_n \\
@VVV @VVV \\
1_i @>>> \coprod 1.
\end{CD}$
Pullbacks preserve coproducts, so the result follows from the fact that in a topos different inclusions in a coproduct don't intersect.