# a x 1 \cong a

A quick note: $A \times 1 \cong A$ where $1$ is the terminal object.

Consider the following diagram,



we have $id_a = \pi_1 \circ \langle id_a, ! \rangle$ already, and we attempt to find whether $\langle id_a, ! \rangle \circ \pi_1$ is equal to $id_{a \times a}$. However, $\langle id_a, ! \rangle \circ \pi_1 = \langle id_a \circ \pi_1, ! \circ \pi_1 \rangle$ where

.

Since $1$ is a terminal object, the morphism $a \times a \rightarrow 1$ is unique. So, $! \circ \pi_1 = \pi_2$ and therefore $\langle id_a \circ \pi_1, ! \circ \pi_1 \rangle = \langle \pi_1, \pi_2 \rangle$ which is the identity of $a \times 1$.

To see that $\langle \pi_1, \pi_2 \rangle$ is the identity, we just draw the following diagram,



where $id_{a \times b}$ also makes the diagram commute. Since $\langle \pi_1, \pi_2 \rangle$ is unique, it must be as same as $id_{a \times b}$.

Also, to see $\langle f, g \rangle \circ h = \langle f \circ h, g \circ h\rangle$ we just draw the following diagram,



where $\langle f, g \rangle \circ h = \langle f \circ h , g \circ h \rangle$ by the similar argument.