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.

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