[proofplan] We use the assumed left-identity property of $e$ on the single morphism $\operatorname{id}_A: A \to A$. Since $\operatorname{id}_A$ has codomain $A$, it is one of the morphisms $g \in \mathcal C(A,A)$ covered by the second hypothesis with $B = A$. Substituting this morphism gives $e \circ \operatorname{id}_A = \operatorname{id}_A$, while the category identity law gives $e \circ \operatorname{id}_A = e$, so the two morphisms are equal. [/proofplan] [step:Apply the assumed identity law for $e$ to $\operatorname{id}_A$] Take $B := A$. The identity morphism of $A$ in the category $\mathcal C$ is a morphism \begin{align*} \operatorname{id}_A \in \mathcal C(A,A). \end{align*} Using the second hypothesis with $g := \operatorname{id}_A$, we obtain \begin{align*} e \circ \operatorname{id}_A = \operatorname{id}_A. \end{align*} [/step] [step:Use the category identity law to identify the same composite with $e$] Since $e \in \mathcal C(A,A)$ and $\operatorname{id}_A$ is the identity morphism on $A$, the identity axiom in the category $\mathcal C$ gives \begin{align*} e \circ \operatorname{id}_A = e. \end{align*} Combining this equality with the equality obtained from the hypothesis gives \begin{align*} e = e \circ \operatorname{id}_A = \operatorname{id}_A. \end{align*} Thus $e = \operatorname{id}_A$, as required. [/step]