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
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations