Uniqueness of Left Adjoint

Question

On nLab, it says that the left adjoint of a functor is unique, but it does not give a proof. Most of the proofs I have seen use the Yoneda lemma, but the book I am using states this fact (without proof) before stating the Yoneda lemma. How is this fact proven?

Answer 1

The left adjoint is unique up to natural isomorphism. To see this, let $G : \mathcal{D} \to \mathcal{C}$ be a functor and suppose $F,F' : \mathcal{C} \to \mathcal{D}$ with are two left adjoints for $G$. Let $(\eta,\varepsilon)$ and $(\eta',\varepsilon')$ be the units and counits of the adjunctions $F \dashv G$ and $F' \dashv G$, respectively.

Define natural transformations $\alpha : F \to F'$ and $\beta : F' \to F$ by $$\alpha = \varepsilon_{F'} \circ F\eta' : F \xrightarrow{F\eta'} FGF' \xrightarrow{\varepsilon_{F'}} F'$$ $$\beta = \varepsilon'_F \circ F'\eta : F' \xrightarrow{F'\eta} F'GF \xrightarrow{\varepsilon'_F} F$$

We now verify that $\alpha$ and $\beta$ are mutually inverse. Consider the following diagram:

\begin{matrix} && FGF && \xrightarrow{=} && FGF && \\ & {\scriptsize F\eta} \nearrow && \searrow{\scriptsize F\eta'_{GF}} && {\scriptsize FG\varepsilon'_F}\nearrow && \searrow {\scriptsize \varepsilon_F} & \\ F && {\scriptsize \text{nat. } \eta'} && FGF'GF && {\scriptsize \text{nat. } \varepsilon} && F \\ & {\scriptsize F\eta'}\searrow && \nearrow{\scriptsize FGF'\eta} && {\scriptsize \varepsilon_{F'GF}} \searrow && \nearrow{\scriptsize \varepsilon'_F} & \\ && FGF' && {\scriptsize \text{nat. } \varepsilon} && F'GF && \\ & && \searrow{\scriptsize \varepsilon_{F'}} && {\scriptsize F'\eta} \nearrow && & \\ && && F' && && \\ \end{matrix} Analysing this, we see that:

• The composite from $F$ to $F$ along the bottom is $\beta \circ \alpha$;
• The left, middle-bottom and right squares commute by naturality of the various units and counits as indicated in the diagram;
• The top triangle commutes by the triangle identity for $F' \dashv G$.

Hence the whole diagram commutes.

Finally, composite from $F$ to $F$ along the top is $\mathrm{id}_F$ by the triangle identity for $F \dashv G$. It follows that $\beta \circ \alpha = \mathrm{id}_F$.

A similar diagram proves $\alpha \circ \beta = \mathrm{id}_{F'}$, so that $F \cong F'$.

A dual argument proves uniqueness (up to natural isomorphism) of right adjoints.