Image of Canonical Injection is Kernel of Projection

Theorem
Let $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$ be groups with identity elements $e_1$ and $e_2$ respectively.

Let $\left({G_1 \times G_2, \circ}\right)$ be the group direct product of $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$

Let:
 * $\operatorname{pr}_1: \left({G_1 \times G_2, \circ}\right) \to \left({G_1, \circ_1}\right)$ be the first projection from $\left({G_1 \times G_2, \circ}\right)$ to $\left({G_1, \circ_1}\right)$
 * $\operatorname{pr}_2: \left({G_1 \times G_2, \circ}\right) \to \left({G_2, \circ_2}\right)$ be the second projection from $\left({G_1 \times G_2, \circ}\right)$ to $\left({G_2, \circ_2}\right)$.

Let:
 * $\operatorname{in}_1: \left({G_1, \circ_1}\right) \to \left({G_1 \times G_2, \circ}\right)$ be the canonical injection from $\left({G_1, \circ_1}\right)$ to $\left({G_1 \times G_2, \circ}\right)$


 * $\operatorname{in}_2: \left({G_2, \circ_2}\right) \to \left({G_1 \times G_2, \circ}\right)$ be the canonical injection from $\left({G_2, \circ_2}\right)$ to $\left({G_1 \times G_2, \circ}\right)$.

Then:
 * $(1): \quad \operatorname{Im} \left({\operatorname{in}_1}\right) = \ker \left({\operatorname{pr}_2}\right)$
 * $(2): \quad \operatorname{Im} \left({\operatorname{in}_2}\right) = \ker \left({\operatorname{pr}_1}\right)$

That is:
 * the image of the (first) canonical injection is the kernel of the second projection
 * the image of the (second) canonical injection is the kernel of the first projection.

Proof
The canonical injection $\operatorname{in}_1: \left({G_1, \circ_1}\right) \to \left({G_1 \times G_2, \circ}\right)$ is defined as:
 * $\forall x \in G_1: \operatorname{in}_1 \left({x}\right) = \left({x, e_2}\right)$

Thus:
 * $\operatorname{Im} \left({\operatorname{in}_1}\right) = \left\{{\left({x, e_2}\right): x \in G_1}\right\}$

The second projection $\operatorname{pr}_2: \left({G_1 \times G_2, \circ}\right) \to \left({G_2, \circ_2}\right)$ is defined as:
 * $\forall \left({x, y}\right) \in G_1 \times G_2: \operatorname{pr}_2 \left({x, y}\right) = y$

Thus by definition of kernel:
 * $\ker \left({\operatorname{pr}_2}\right) = \operatorname{pr}_2^{-1} \left({e_2}\right) = \left\{{\left({x, e_2}\right): x \in G_1}\right\}$

As can be seen:
 * $\operatorname{Im} \left({\operatorname{in}_1}\right) = \ker \left({\operatorname{pr}_2}\right)$

Similarly:


 * $\operatorname{Im} \left({\operatorname{in}_2}\right) = \left\{{\left({e_1, y}\right): y \in G_2}\right\}$
 * $\ker \left({\operatorname{pr}_1}\right) = \operatorname{pr}_1^{-1} \left({e_1}\right) = \left\{{\left({e_1, y}\right): y \in G_2}\right\}$

Hence the result.