Michael wrote:
> Is the identity morphism an isomorphism?
Yes. Let's see why! An **isomorphism** is a morphism \\(f \colon x \to y\\) that has a morphism \\(g \colon y \to x\\) for which
\[ gf = 1_x \textrm{ and } fg = 1_y .\]
We call \\(g\\) an **inverse** of \\(f\\).
Is an identity morphism \\(1_x \colon x \to x \\) an isomorphism? Yes: it has itself as its own inverse! In this example \\(y = x\\), \\(f = 1_x\\), \\(g = 1_x\\), and
\[ 1_x 1_x = 1_x \textrm{ and } 1_x 1_x = 1_x .\]
So the inverse of an identity morphism is itself.
The same argument shows that an identity natural transformation is a natural isomorphism.
> Is the identity morphism an isomorphism?
Yes. Let's see why! An **isomorphism** is a morphism \\(f \colon x \to y\\) that has a morphism \\(g \colon y \to x\\) for which
\[ gf = 1_x \textrm{ and } fg = 1_y .\]
We call \\(g\\) an **inverse** of \\(f\\).
Is an identity morphism \\(1_x \colon x \to x \\) an isomorphism? Yes: it has itself as its own inverse! In this example \\(y = x\\), \\(f = 1_x\\), \\(g = 1_x\\), and
\[ 1_x 1_x = 1_x \textrm{ and } 1_x 1_x = 1_x .\]
So the inverse of an identity morphism is itself.
The same argument shows that an identity natural transformation is a natural isomorphism.
Version 2.1.8p2
