Properties of the Projective Limit (in $\textbf{Set}$)

Question

Let $I$ be an index set with a preorder $\leq$ and let $(G_i)_{i \in I}$ be a family of sets. Furthermore, for all $i,j \in I$ with $i \leq j$ let $f_{ij} \colon G_i \longrightarrow G_j $ be maps such that (i) $f_{ii} = \operatorname{id}_{G_i}$ for all $i \in I$ and (ii) $f_{jk} \circ f_{ij} = f_{ik}$ for all $i \leq j \leq k$.

Then there exists a set $G$ called a projective limit (of the projective system above) together with projections $\pi_i \colon G \longrightarrow G_i $ for all $i \in I$ such that for all $i,j \in I$ with $i \leq j$ we have $f_{ij} \circ \pi_i = \pi_j$ and that the following universal property holds:

Given any set $H$ together with maps $p_i \colon H \longrightarrow G_i $ for all $i \in I$ such that for all $i,j \in I$ with $i \leq j$ we have $f_{ij} \circ p_i = p_j$, there exists a unique map $u \colon H \longrightarrow G$ such that $\pi_i \circ u = p_i$ for all $i \in I$.

(The universal property implies that the projective limit is unique up to unique bijections.) Usually one proofs the existence of a projective limit (in $\textbf{Set}$) by showing that the set $$ G := \left\{ (g_i)_{i \in I} \in \prod_{i \in I} G_i \;\middle|\; f_{ij}(g_i) = g_j \text{ for all $i \leq j$}\right\} $$ together with the maps $$ \begin{align*}\pi_i \colon\ \qquad G &\longrightarrow G_i \\ (g_j)_{j \in I} &\longmapsto g_i \end{align*}$$ has all the claimed properties.

Now, knowing this explicit construction, it is clear that (for any projective limit $G$ in $\textbf{Set}$):

1. The maps $\require{enclose}\enclose{horizontalstrike}{\pi_i \colon G \longrightarrow G_i}$ are surjective for all $\enclose{horizontalstrike}{i \in I}$.
2. Given $x,y \in G$ such that $\pi_i(x) = \pi_i(y)$ for all $i \in I$, we can conclude $x = y$.

My question: Since the universal property is said to characterize the projective limit, I was wondering, is it somehow possible to derive these two properties (1) and (2) directly from it, i.e. without referring back to the explicit construction?

Answer 1

Property 1. is not correct* (so I will better not prove it). Property 2. follows since $x,y$ correspond to maps $x,y : 1 \to G$ (where $1$ is the terminal set) with the same result when composing with any $\pi_i : G \to G_i$, hence they are the same by the universal property. Actually, using $\hom(1,-)$ you can immediately deduce the element structure from the universal property.

*Take $I = \{0,1\}$ with no relation except $0 \leq 0$, $1 \leq 1$, so a projective limit is just the product $G_0 \times G_1$. When $G_0 = \emptyset$ and $G_1 \neq \emptyset$, then $\pi_1 : G_0 \times G_1 \to G_1$ is not surjective. Maybe you want to assume that $I=(\mathbb{N},\geq)$ and that all $G_i \to G_j$ are surjective?