Naturality of the counit of an adjunction

Question

Consider the picture below which is from my lecture notes.

I do not understand how to prove the naturality of counit. I tried to use the naturality of the Hom functor but did not get anywhere. Any help would be appreciated!

Answer 1

Naturality of $\epsilon$ means that for every $f: d_1 \rightarrow d_2$ the diagram

commutes. This follows immediately from the diagram below

where the two squares commute by definition of the adjunction. (I denote with $f_*$ concatenation with $f$ and with $f^*$ "pre-concatenation" with $f$.)

To be more precise, start to consider the upper square in the diagram. On the left hand side, $\epsilon_{d_1}$ is mapped by $f_*$ to $\epsilon_{d_1} \circ f$. On the right hand side, $\mathrm{id}_{Rd_1}$ is mapped by $(Rf)_*$ to $Rf$. Since the isomorphisms from the adjunction are natural in the second component, the isomorphism $\mathrm{Hom}_C(LRd_1, d_2) \cong \mathrm{Hom}_D(Rd_1, Rd_2)$ maps $\epsilon_{d_1} \circ f$ to $Rf$.

The same argument applied to the square below yields that the same isomorphism maps $LRf \circ \epsilon_{d_2}$ to $Rf$. But then we obtain $$ LRf \circ \epsilon_{d_2} = \epsilon_{d_1} \circ f, $$ which is exactly what we wanted to show.

Answer 2

By definition, a family of morphisms $\alpha_X\colon FX\to GX$ is a natural transformation from a functor $F$ to a functor $G$ if $\alpha_Y\circ Fh=Gh\circ\alpha_X$ for any $h\colon X\to Y$. But a family of morphisms $\alpha_X\colon FX\to GX$ is equivalently a family of elements $\alpha_X\in\mathrm{Hom}(FX,GX)$, and the naturality condition translates to an "extra-naturality" condition that $\mathrm{Hom}(Fh,GY)(\alpha_Y)=\mathrm{Hom}(FX,Gh)(\alpha_X)$.

In particular, the identity natural transformation $\mathrm{id}_{FX}\colon FX\to FX$ corresponds to an extra-natural family of identity elements $\mathrm{id}_{FX}\in\mathrm{Hom}(FX,FX)$, and the counit family of elements $\epsilon_X\in\mathrm{Hom}(GFX,X)$ is the image of this family of elements under the natural isomorphism $\mathrm{Hom}(FX,FY)\cong\mathrm{Hom}(GFX,Y)$. Thus what is needed is to show that the extranaturality of the family of identity elements is preserved by the above natural isomorphisms.

It is straightforward to check this by hand: if $\alpha_{X,Y}\colon D(X,Y)\to E(X,Y)$ is a natural transformation, and $d_X\in D(X,X)$ is an extra-natural family of elements, then $E(X,h)\circ\alpha_{X,X}(d_X)=\alpha_{X,Y}\circ D(X,h)(d_X)=\alpha_{X,Y}\circ D(h,X)(d_Y)=E(h,Y)\circ\alpha_{Y,Y}(d_Y)$.

Another way of concluding that extranatural families of elements are preserved by natural transformation is to use a variation of the Yoneda lemma which asserts that extranatural families of elements are in bijective correspondnce with natural transformation from $\mathrm{Hom}(-,-)$. Indeed, a natural transformation $d\colon\mathrm{Hom}(-,-)\to D(-,-)$ consists of a family of morphisms $d_{X,Y}\colon\mathrm{Hom}(X,Y)\to D(X,Y)$ such that for $f\colon X_1\to X_2$ and $g\colon Y_1\to Y_2$ we have $d_{X_1,Y_2}\circ \mathrm{Hom}(f,g)=D(f,g)\circ d_{X_2,Y_1}$.

Since $\mathrm{Hom}(h,Y)(\mathrm{id}_Y)=h=\mathrm{Hom}(X,h)(\mathrm{id_X})(h)$ for each $h\in\mathrm{Hom}(X,Y)$, it follows that $D(X,h)\circ d_{X,X}(\mathrm{id_X})=d_{X,Y}\circ\mathrm{Hom}(X,h)(\mathrm{id_X})=d_{X,Y}(h)=d_{X,Y}\circ\mathrm{Hom}(h,Y)(\mathrm{id_Y})=D(h,Y)\circ d_{Y,Y}(\mathrm{id_Y})$.

In other words, each natural transformation $d\colon\mathrm{Hom}(-,-)\to D(-,-)$ determines an extra-natural family of elements $d_X\in D(X,X)$. The variation of the Yoneda lemma now asserts that the inverse transformation given by $d_{X,Y}(h)=D(h,Y)(d_Y)=D(X,h)(d_X)$ is natural. Indeed, one the one hand $d_{X_1,Y_2}\circ\mathrm{Hom}(f,g)(h)=d_{X_1,Y_2}(ghf)$.

On the other hand, $D(f,g)\circ d_{X_2,Y_1}(h)=D(f,g)\circ D(X_2,h)(d_{X_2})=D(X_2,g)\circ D(f,Y_1)\circ D(X_2,h)(d_{X_2})$ which further equals $D(X_2,g)\circ D(X_2,h)\circ D(f,Y_1) (d_{X_2})=D(X_2,g)\circ D(X_2,h)\circ D(X_1,f)(d_{X_1})=D(X_2,ghf)(d_{X_1})=d_{X_1,X_2}(ghf)$.