Adjoint Functors: Unit is Right Invertible? $\DeclareMathOperator{\Hom}{Hom}$
Suppose there is an adjunction $F: C \to D$, $G: D \to C$, $F \dashv G$. Let $\eta: 1_C \implies GF$ be the unit of the adjunction.
Suppose there is a natural transformation, $\kappa: F \implies F$. Define the transpose of the natural transformation to be $G\kappa$ (which is computed as $(G\kappa)_x \equiv G(\kappa_x)$.)
Now the claim (in "Category Theory in Context", by Emily Riehl, Section 4.4.1, "calculus of adjunctions"), where $\kappa \sim \theta' \circ \theta$, is that to prove $\kappa = 1_F$, it suffices to prove $(G\kappa) \circ \eta = (G1_F) \circ \eta$, ie, $(G\kappa) \circ \eta = 1_{GF} \circ \eta = \eta$.
I see no reason why the condition that $(G\kappa) \circ \eta = \eta$ is sufficient to prove that $\kappa = 1_F$, since there is no guarantee that $\eta$ is right-invertible.
The triangle identities only tell us that $1_F = \epsilon F \circ F \eta$ and $1_G = G \epsilon \circ \eta G$.
What am I missing?
Category Theory in Context, Sec 4.4.1, "Calculus of Adjunctions"

 A: 
I see no reason why the condition that $(G\kappa) \circ \eta = \eta$ is sufficient to prove that $\kappa = 1_F$.

The unit $\eta$ of an adjunction is an universal arrow.
This means that for every object $x$, the morphism $\eta_x:x\to(G\circ F)(x)$ is an initial object in the comma category $x\downarrow G$.
Adjoint functors can be defined in terms of universal arrows (see here).

More explicity: if $F \dashv G$ is an adjunction with unit $\eta$, then for every object $x$ of $C$ the morphism $\eta_x:x\to G(F(x))$ has the following universal property: for every object $y$ of $D$ and every morphism $f:x\to G(y)$ there exists one and only one morphism $h:F(x)\to y$ in $D$ such that $f=G(h)\circ\eta_x$.
To answer your question: let $\kappa$ be a natural transformation such that $G(\kappa)\circ\eta=\eta$.
Then for every object $x$ of $C$ we have $G(\kappa_x)\circ\eta_x=\eta_x=G(1_{F(x)})\circ\eta_x$.
By uniqueness of morphism $h$ above, we can conclude $\kappa_x=1_{F(x)}$, that's $\kappa=1_F$.
