Mappings in Product of Sets are Surjections/Family of Sets

Theorem
Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ be an indexed family of sets.

Let $\left({P, \left\langle{\phi_i}\right\rangle_{i \in I}}\right)$ be a product of $S$ and $T$.

Then for all $i \in I$, $\phi_i$ is a surjection.

Proof
From the definition:
 * For all sets $X$ and all indexed families $\left\langle{f_i}\right\rangle_{i \mathop \in I}$ of mappings $f_i: X \to S_i$ there exists a unique mapping $h: X \to P$ such that:
 * $\forall i \in I: \phi_i \circ h = f_i$

Let:
 * $i \in I$
 * $X = S_i$
 * $f_i = I_{S_i}$

where $I_{S_i}$ is the identity mapping on $S_i$.

Then we have:
 * $\phi_i \circ h = I_{S_i}$

From Identity Mapping is Surjection:
 * $I_{S_i}$ is a surjection.

From Surjection if Composite is Surjection:
 * $\phi_i$ is a surjection.

$i \in I$ is arbitrary so the argument applies whatever its value.

Hence the result.