Projections on Direct Product of Group Homomorphisms

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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:

$\operatorname{pr}_1: H_1 \times H_2 \to H_1$ be the first projection from $H_1 \times H_2$ to $H_1$
$\operatorname{pr}_2: H_1 \times H_2 \to H_2$ be the second projection from $H_1 \times H_2$ to $H_2$.


Then:

$(1) \quad \operatorname{pr}_1 \circ \left({f_1 \times f_2}\right) = f_1$
$(2) \quad \operatorname{pr}_2 \circ \left({f_1 \times f_2}\right) = f_2$

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 $\operatorname{pr}_1 \circ \left({f_1 \times f_2}\right)$ 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 $\operatorname{pr}_1 \circ \left({f_1 \times f_2}\right)$ and $f_1$ are the same.

$\blacksquare$


Matching Codomains

From Codomain of Composite Relation, the codomain of $\operatorname{pr}_1 \circ \left({f_1 \times f_2}\right)$ is the codomain of $\operatorname{pr}_1$.

By definition of the first projection $H_1 \times H_2$, the codomain of $\operatorname{pr}_1$ is $H_1$.

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


Thus the codomains of $\operatorname{pr}_1 \circ \left({f_1 \times f_2}\right)$ and $f_1$ are the same.

$\blacksquare$


Matching Images

Let $g \in G$.

Then:

\(\displaystyle \operatorname{pr}_1 \circ \left({f_1 \times f_2}\right) \left({g}\right)\) \(=\) \(\displaystyle \operatorname{pr}_1 \left({\left({f_1 \times f_2}\right) \left({g}\right)}\right)\) Definition of Composition of Mappings
\(\displaystyle \) \(=\) \(\displaystyle \operatorname{pr}_1 \left({f_1 \left({g}\right), f_2 \left({g}\right)}\right)\) Definition of Direct Product of $f_1$ and $f_2$
\(\displaystyle \) \(=\) \(\displaystyle f_1 \left({g}\right)\) Definition of First Projection

So:

$g \in G \implies \operatorname{pr}_1 \circ \left({f_1 \times f_2}\right) \left({g}\right) = f_1 \left({g}\right)$

and so Equality of Mappings has been demonstrated.

$\blacksquare$


Sources