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.