[proofplan] The proof uses only the identity laws and associativity of composition in the category $\mathcal C$. Since $h$ is a right inverse of $f$, the identity morphism on $B$ can be replaced by $f \circ h$. Associativity then moves the parentheses so that the left inverse equation for $g$ can be applied, reducing $g$ to $h$. [/proofplan] [step:Insert the identity on $B$ after $g$ and replace it using $h$] Since $g: B \to A$, the right identity law in $\mathcal C$ gives \begin{align*} g = g \circ \operatorname{id}_B. \end{align*} Because $h$ is an inverse of $f$, we have $f \circ h = \operatorname{id}_B$. Substituting this identity into the previous equality gives \begin{align*} g = g \circ (f \circ h). \end{align*} [/step] [step:Reassociate the composition and use that $g$ is a left inverse of $f$] The morphisms $g: B \to A$, $f: A \to B$, and $h: B \to A$ are composable in the displayed order, so associativity of composition in $\mathcal C$ gives \begin{align*} g \circ (f \circ h) = (g \circ f) \circ h. \end{align*} Because $g$ is an inverse of $f$, we have $g \circ f = \operatorname{id}_A$. Therefore \begin{align*} g = (g \circ f) \circ h = \operatorname{id}_A \circ h. \end{align*} [/step] [step:Apply the left identity law to conclude equality of the inverses] Since $h: B \to A$, the left identity law in $\mathcal C$ gives \begin{align*} \operatorname{id}_A \circ h = h. \end{align*} Combining the preceding equalities yields \begin{align*} g = h. \end{align*} Thus any two inverse morphisms of $f$ coincide. [/step]