It was quite encouraging to learn in comments to previous lecture that "elementary" proofs are hard for everybody, so let me try

**Puzzle 140**: Prove that any morphism has at most one inverse.

Let \\(g: y \to x\\) to be an inverse of \\(f: x \to y\\), and \\(h: y \to x\\) to be another inverse of \\(f\\). In this case we have the right inverse condition hold for both:

\[ f \circ g = 1_y \]

\[ f \circ h = 1_y \]

This implies that

\[ f \circ g = f \circ h \]

Composing both sides with \\(g\\) gives

\[ g \circ f \circ g = g \circ f \circ h \]

Using left inverse condition, \\(g \circ f = 1_x\\), we get

\[1_x \circ g = 1_x \circ h\]

\[ g = h \]

So all inverses of a given function are equal (isomorphic?) to each other, so we can say that such inverse is unique (up to isomorphism?).

It also seems that having left and right inverses together form a universal property (I'm still not quite sure what are these exactly), which uniquely identifies the inverse of a given function.

**Puzzle 140**: Prove that any morphism has at most one inverse.

Let \\(g: y \to x\\) to be an inverse of \\(f: x \to y\\), and \\(h: y \to x\\) to be another inverse of \\(f\\). In this case we have the right inverse condition hold for both:

\[ f \circ g = 1_y \]

\[ f \circ h = 1_y \]

This implies that

\[ f \circ g = f \circ h \]

Composing both sides with \\(g\\) gives

\[ g \circ f \circ g = g \circ f \circ h \]

Using left inverse condition, \\(g \circ f = 1_x\\), we get

\[1_x \circ g = 1_x \circ h\]

\[ g = h \]

So all inverses of a given function are equal (isomorphic?) to each other, so we can say that such inverse is unique (up to isomorphism?).

It also seems that having left and right inverses together form a universal property (I'm still not quite sure what are these exactly), which uniquely identifies the inverse of a given function.