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It is well known that there are (at least) $4$ equivalent characterizations of an adjunction:

  1. An antiparallel pair of functors $F:\mathcal{C}\rightleftarrows\mathcal{D}:G$ together with a natural transformation $\eta:1_\mathcal{C}\Rightarrow G\circ F$ satisfying an appropriate universal property.
  2. An antiparallel pair of functors $F:\mathcal{C}\rightleftarrows\mathcal{D}:G$ together with a natural transformation $\epsilon:F\circ G\Rightarrow 1_\mathcal{D}$ satisfying an appropriate universal property.
  3. An antiparallel pair of functors $F:\mathcal{C}\rightleftarrows\mathcal{D}:G$ together with functions $f_{C,D}:{\bf Hom}_\mathcal{C}(C,G(D))\leftrightarrows{\bf Hom}_\mathcal{D}(F(C),D):g_{C,D}$ which are natural in $C$ and $D$, in the appropriate sense.
  4. An antiparallel pair of functors $F:\mathcal{C}\rightleftarrows\mathcal{D}:G$ together with natural transformations $\eta:1_\mathcal{C}\Rightarrow G\circ F$ and $\epsilon:F\circ G\Rightarrow1_\mathcal{D}$ satisfying the triangle identities.

What are some examples of adjunctions that are significantly easier to prove using one characterization over the others?

An example I recently encountered is showing that, given an adjunction $F\dashv G:\mathcal{C}\rightleftarrows\mathcal{D}$ between categories where $\mathcal{C}$ has pullbacks, each slice functor $F_{/X}:\mathcal{C}/X\to\mathcal{D}/F(X)$ has a right adjoint $$G^{/X}:\mathcal{D}/F(X)\to\mathcal{C}/X$$ $$(D,g)\mapsto(X\times_{_{G(F(X))}}G(D),G(g)_{\eta_X})$$ $$h:(D,g)\to(D',g')\longmapsto1_X\times_{_{G(F(X))}}G(h):(X\times_{_{G(F(X))}}G(D),G(g)_{\eta_X})\to (X\times_{_{G(F(X))}}G(D'),G(g')_{\eta_X})$$ where $G(g)_{\eta_X}$ denotes the pullback of $G(g)$ along $\eta_X$. This is relatively simple to prove from the counit side ($2.$), but I had originally started to prove it from the unit side ($1.$) and stopped at the universal property part of the proof because it seemed ugly.

In this situation $F_{/X}$ is a simpler functor than $G^{/X}$, so it makes sense that the universal property part of the proof is easier using $(2.)$ since we need to find an arrow and look at it's image under $F_{/X}$, as opposed to looking at the image of an arrow under $G^{/X}$ for the universal property part of ($1.$). I haven't tried ($3.$) or ($4.$) in this situation because they require more of an 'arrow hunt', and suspect that ($2.$) is optimal for proof ease.

The general heuristic here suggests that we might find examples of the phenomenon in question by looking at adjoint functor pairs where one functor is significantly simpler than the other, e.g. ${\sf Free}\dashv{\sf Forgetful}$ adjunctions and the like. All examples are appreciated.

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