# Projections on Direct Product of Group Homomorphisms

## Theorem

Let $G, H_1$ and $H_2$ be groups.

Let $H_1 \times H_2$ be the group direct product of $H_1$ and $H_2$.

Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be group homomorphisms.

Let $f_1 \times f_2: g \to H_1 \times H_2$ be the direct product of $f_1$ and $f_2$.

Let:

$\pr_1: H_1 \times H_2 \to H_1$ be the first projection from $H_1 \times H_2$ to $H_1$
$\pr_2: H_1 \times H_2 \to H_2$ be the second projection from $H_1 \times H_2$ to $H_2$.

Then:

$(1) \quad \pr_1 \circ \paren {f_1 \times f_2} = f_1$
$(2) \quad \pr_2 \circ \paren {f_1 \times f_2} = f_2$

where $\circ$ is the operation of composition of mappings.

## Proof

The proof of $(2)$ follows exactly the same lines as the proof of $(1)$, so the proof of $(1)$ only will be shown here.

### Matching Domains

From Domain of Composite Relation, the domain of $\pr_1 \circ \paren {f_1 \times f_2}$ is the domain of $f_1 \times f_2$.

By definition of the direct product of $f_1$ and $f_2$, the domain of $f_1 \times f_2$ is $G$.

Also by definition, the domain of $f_1$ is also $G$.

Thus the domains of $\pr_1 \circ \paren {f_1 \times f_2}$ and $f_1$ are the same.

$\blacksquare$

### Matching Codomains

From Codomain of Composite Relation, the codomain of $\pr_1 \circ \paren {f_1 \times f_2}$ is the codomain of $\pr_1$.

By definition of the first projection $H_1 \times H_2$, the codomain of $\pr_1$ is $H_1$.

Also by definition, the codomain of $f_1$ is also $H_1$.

Thus the codomains of $\pr_1 \circ \paren {f_1 \times f_2}$ and $f_1$ are the same.

$\blacksquare$

### Matching Images

Let $g \in G$.

Then:

 $\ds \map {\pr_1 \circ \paren {f_1 \times f_2} } g$ $=$ $\ds \map {\pr_1} {\map {\paren {f_1 \times f_2} } g}$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map {\pr_1} {\map {f_1} g, \map {f_2} g}$ Definition of Direct Product of $f_1$ and $f_2$ $\ds$ $=$ $\ds \map {f_1} g$ Definition of First Projection

So:

$g \in G \implies \map {\pr_1 \circ \paren {f_1 \times f_2} } g = \map {f_1} g$

and so Equality of Mappings has been demonstrated.

$\blacksquare$