Mappings in Product of Sets are Surjections

Theorem
Let $S$ and $T$ be sets.

Let $\left({P, \phi_1, \phi_2}\right)$ be a product of $S$ and $T$.

Then $\phi_1$ and $\phi_1$ are surjections.

Proof
From the definition:
 * For all sets $X$ and all mappings $f_1: X \to S$ and $f_2: X \to T$ there exists a unique mapping $h: X \to P$ such that:
 * $\phi_1 \circ h = f_1$
 * $\phi_2 \circ h = f_2$

Let $X = S$ and let $f_1 = I_S$ where $I_S$ is the identity mapping on $S$.

Then we have:
 * $\phi_1 \circ h = I_S$

We have from Identity Mapping is Surjection that $I_S$ is a surjection.

From Surjection if Composite is a Surjection it follows that $\phi_1$ is a surjection.

Similarly, let $X = T$ and let $f_2 = I_T$ where $I_T$ is the identity mapping on $T$.

Then we have:
 * $\phi_2 \circ h = I_T$

We have from Identity Mapping is Surjection that $I_T$ is a surjection.

From Surjection if Composite is a Surjection it follows that $\phi_2$ is a surjection.