Mappings in Product of Sets are Surjections/Family of Sets

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Theorem

Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.

Let $\struct {P, \family {\phi_i} _{i \mathop \in I} }$ 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 $\family {f_i}_{i \mathop \in I}$ 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.

$\blacksquare$


Sources