Natural transformations of Hom-sets “transport” natural transformations from one pair of functors to another? (Reference)

Question

Question 1: Does anyone know a name, or have a reference, for the following lemma?

$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\F}{\mathscr{F}}$$\newcommand{\G}{\mathscr{G}}$$\newcommand{\op}{\operatorname{op}}$$\newcommand{\C}{\mathscr{C}}$ $\newcommand{\Id}{\operatorname{Id}}$$\newcommand{\Ob}{\operatorname{Ob}}$$\newcommand{\Set}{\operatorname{Set}}$$\newcommand{\eval}{\operatorname{eval}}$Lemma: Given functors $G_1,G_2: \C \to \G$ such that there exists a natural transformation $G_1 \implies G_2$, and functors $F_1, F_2: \C \to \F$ such that there exists a natural transformation ${ \Hom_{\G} \circ (G_1^{\op}, G_2) \implies \Hom_{\F} \circ (F_1^{\op}, F_2) }$, then there exists a natural transformation $F_1 \implies F_2$.

Question 2: If the above lemma can be used to prove the Yoneda lemma, how would one do so? If the above lemma is a corollary of the Yoneda lemma, then how?

Optional context: I will put a proof of this lemma in an answer below.

I found the lemma when trying to prove that the natural bijection of Hom sets in the definition of an adjunction implies the existence of unit and counit natural transformations. I was confused because the proof seemed to use very little of the assumptions available, and what I was able to distill the proof to was the above lemma.

Superficially, it seems like this lemma should be related to the Yoneda lemma. Both lemmas involve natural transformations where at least one of the functors is a Hom functor, and both can be used to prove the existence of the unit and counit natural transformations in an adjunction. They also both seem to have proofs involving natural transformations being determined by the components of another natural transformation (given that identity morphisms are the components of identity natural transformations).

If this lemma can be used to prove Yoneda, (1) it seems odd that this lemma doesn't have a name / isn't more widely known, and (2) I haven't been able to figure out how to use it to prove Yoneda.

Answer 1

$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\F}{\mathscr{F}}$$\newcommand{\G}{\mathscr{G}}$$\newcommand{\op}{\operatorname{op}}$$\newcommand{\C}{\mathscr{C}}$ $\newcommand{\Id}{\operatorname{Id}}$$\newcommand{\Ob}{\operatorname{Ob}}$Lemma: Given functors $G_1,G_2: \C \to \G$ such that there exists a natural transformation $G_1 \implies G_2$, and functors $F_1, F_2: \C \to \F$ such that there exists a natural transformation ${ \Hom_{\G} \circ (G_1^{\op}, G_2) \implies \Hom_{\F} \circ (F_1^{\op}, F_2) }$, then there exists a natural transformation $F_1 \implies F_2$.

Proof: For every $c \in \Ob(\C)$, let $\lambda_c$ denote the corresponding component of the natural transformation $G_1 \implies G_2$.

For every ${(c_1, c_2) \in \Ob(\C^{\op} \times \C)}$, let $\eta_{c_1, c_2}$ denote the corresponding component of the natural transformation ${ \Hom_{\G} \circ (G_1^{\op}, G_2) \implies \Hom_{\F} \circ (F_1^{\op}, F_2) }$.

Then the claim is that the $\eta_{c,c}(\lambda_c) =: \mu_c$ for every $c \in \Ob(\C)$ define the components of a natural transformation $F_1 \implies F_2$. What we need to show to prove the claim is that for every $c_1 \overset{h}{\longrightarrow} c_2$ in $\operatorname{Mor}(\C)$ the square:

$$ \require{AMScd} \begin{CD} F_1(c_1) @>> F_1(h) > F_1(c_2) \\ @VV \displaystyle \mu_{c_1} V @VV \displaystyle \mu_{c_2} V \\ F_2(c_1) @>> F_2(h) > F_2(c_2) \end{CD} $$

commutes, i.e. that $\eta_{c_2, c_2}(\lambda_{c_2}) \circ F_1(h) = \mu_{c_2} \circ F_1(h) = F_2(h) \circ \mu_{c_1} = F_2(h) \circ \eta_{c_1, c_1} (\lambda_{c_1})$.

To show this, we look at two naturality squares for $\Hom_{\G} \circ (G_1^{\op}, G_2) \implies \Hom_{\F} \circ (F_1^{\op}, F_2)$:

$$ \require{AMScd} \begin{CD} \Hom_{\G}(G_1(c_1), G_2(c_1)) @> \displaystyle (\Hom_{\G} \circ (G_1^{\op}, G_2))(h_1) > \alpha_1 \mapsto G_2(\pi_2(h_1)) \circ \alpha_1 \circ G_1(\pi_1(h_1)) > \Hom_{\G}(G_1(c_1), G_2(c_2)) @< \displaystyle (\Hom_{\G} \circ (G_1^{\op}, G_2))(h_2) < \alpha_2 \mapsto G_2(\pi_2(h_2)) \circ \alpha_2 \circ G_1(\pi_1(h_2)) < \Hom_{\G}(G_1(c_2), G_2(c_2)) \\ @VV \displaystyle \eta_{c_1, c_1} V @VV \displaystyle \eta_{c_1, c_2} V @VV \displaystyle \eta_{c_2, c_2} V \\ \Hom_{\F}(F_1(c_1), F_2(c_1)) @> \beta_1 \mapsto F_2(\pi_2(h_1)) \circ \beta_1 \circ F_1(\pi_1(h_1)) > \displaystyle (\Hom_{\F} \circ (F_1^{\op}, F_2))(h_1) > \Hom_{\F}(F_1(c_1), F_2(c_2)) @< \beta_2 \mapsto F_2(\pi_2(h_2)) \circ \beta_2 \circ F_1(\pi_1(h_2)) < \displaystyle (\Hom_{\F} \circ (F_1^{\op}, F_2))(h_2) < \Hom_{\F} (F_1(c_2), F_2(c_2)) \end{CD} $$

The crux then comes down to making clever choices for the morphisms ${(c_1, c_1) \overset{h_1}{\longrightarrow} (c_1, c_2)}$ and ${(c_2, c_2) \overset{h_2}{\longrightarrow} (c_1, c_2)}$ in ${\C^{\op} \times \C }$. What works is $h_1 := {(c_1 \overset{\Id_{c_1}}{\longleftarrow} c_1, c_1 \overset{h}{\longrightarrow} c_2 )}$ and $h_2 := {(c_2 \overset{h}{\longleftarrow} c_1, c_2 \overset{\Id_{c_2}}{\longrightarrow} c_2)}$.

Choosing $\lambda_{c_1} \in \Hom_{\G}(G_1(c_1), G_2(c_1))$ and chasing it around the left naturality square leads to $$\eta_{c_1, c_2} (G_2(h) \circ \lambda_{c_1}) = F_2(h) \circ \mu_{c_1}. $$ Choosing $\lambda_{c_2} \in \Hom_{\G}(G_1(c_2), G_2(c_2))$ and chasing it around the right naturality square leads to $$\eta_{c_1, c_2}(\lambda_{c_2} \circ G_1(h)) = \mu_{c_2} \circ F_1(h).$$

Now it may initially seem that all hope is lost because $\eta_{c_1, c_2} (G_2(h) \circ \lambda_{c_1})$ is not obviously equal to $\eta_{c_1, c_2}(\lambda_{c_2} \circ G_1(h))$. However, the $\lambda_c$'s are themselves components of the natural transformation $G_1 \implies G_2$, which associates the morphism $c_1 \overset{h}{\longrightarrow} c_2$ in $\C$ with the naturality square: $$ \require{AMScd} \begin{CD} G_1(c_1) @> G_1(h) >> G_1(c_2) \\ @V \displaystyle \lambda_{c_1} VV @VV \displaystyle \lambda_{c_2} V \\ G_2(c_1) @> G_2(h) >> G_2(c_2) \end{CD} $$ in $\G$. In other words, it is a direct consequence of naturality of $\lambda$ that $G_2(h) \circ \lambda_{c_1} = \lambda_{c_2} \circ G_1(h) =: \tilde{h}$. Hence $$ F_2(h) \circ \mu_{c_1} = \eta_{c_1, c_2} (G_2(h) \circ \lambda_{c_1}) = \eta_{c_1, c_2}(\tilde{h}) = \eta_{c_1, c_2}(\lambda_{c_2} \circ G_1(h)) = \mu_{c_2} \circ F_1(h), $$ with the equality of the first and last term being exactly what was needed to be shown for the $\mu_c$'s to be the components of a natural transformation $F_1 \implies F_2$. $\square$

Proof of existence of unit and counit natural transformations: Let's say we have an adjunction between functors $\newcommand{\B}{\mathscr{B}}\newcommand{\E}{\mathscr{E}}\newcommand{\Set}{\operatorname{Set}}$ $I: \B \to \E$ and $T: \E \to \B$, so a natural isomorphism between the functors $\Hom_{\B} \circ ( (\Id_{\B})^{\op}, T) : \B^{\op} \times \E \to \Set$ and $\Hom_{\E} \circ (I^{\op}, \Id_{\E}): \B^{\op} \times \E \to \Set$.

Via "pre-whiskering" the natural transformation $\Hom_{\E} \circ (I^{\op}, \Id_{\E}) \implies \Hom_{\B} \circ ( (\Id_{\B})^{\op}, T)$ with the functor $((\Id_{\B})^{\op}, I): \B^{\op} \times \B \to \B^{\op} \times \E$, we get a natural transformation $$ \Hom_{\E} \circ (I^{\op}, I) \implies \Hom_{\B} \circ ( (\Id_{\B})^{\op}, T \circ I)$$ of functors $\B^{\op} \times \B \to \Set$. Then the identity natural transformation $\Id_I: I \times I$ combined with the above lemma gives us a natural transformation $\Id_{\B} \implies T \circ I$.

Similarly, via "pre-whiskering" the natural transformation $\Hom_{\B} \circ ( (\Id_{\B})^{\op}, T) \implies \Hom_{\E} \circ (I^{\op}, \Id_{\E})$ with the functor $(T^{\op}, \Id_{\E}): \E^{\op} \times \E \to \B^{\op} \times \E$, we get a natural transformation $$ \Hom_{\B} \circ ( T^{\op}, T) \implies \Hom_{\E} \circ (I^{\op} \circ T^{\op}, \Id_{\E})$$ of functors $\E^{\op} \times \E \to \Set$. Then the identity natural transformation $\Id_T: T \implies T$ combined with the above lemma gives us a natural transformation $I \circ T \implies \Id_{\E}$.

Attempt for question 2:$\newcommand{\eval}{\operatorname{eval}}$ Because this lemma seems simpler and more general than the Yoneda lemma (no restrictions on the categories $\C$, $\F$, or $\G$, only a one-directional natural transformation, not a natural isomorphism), it seems like if there is a relationship to the Yoneda lemma, that the Yoneda lemma would either be a consequence of or a special case of this lemma.

For Yoneda, we are trying to prove a natural isomorphism between two functors ${\C \times [\C, \Set] \to \Set}$, namely $F_1 = \Hom_{[\C, \Set]} \circ ( (\Hom_{\C}(\bullet, -))^{\op}, \Id_{[\C, \Set]})$ (where I have used the non-standard notation $X(\bullet, -)$ to denote currying in the first coordinate) to $F_2 = \eval_{[\C, \Set]}$.

So if Yoneda is a corollary of the above lemma, then we need to find some category $\G$, and two functors $G_1, G_2:\C \times [\C, \Set] \to \G$ such that $G_1$ and $G_2$ are naturally isomorphic and such that there are natural transformations $$\Hom_{\G} \circ (G_1^{\op}, G_2) \implies \Hom_{\Set} ( (Hom_{[\C, \Set]} \circ ( (\Hom_{\C}(\bullet, -))^{\op}, \Id_{[\C, \Set]}))^{\op}, \eval_{[\C, \Set]}) $$ and $$ \Hom_{\G} \circ (G_2^{\op}, G_1) \implies \Hom_{\Set} ( (\eval_{[\C, \Set]})^{\op} , Hom_{[\C, \Set]} \circ ( (\Hom_{\C}(\bullet, -))^{\op}, \Id_{[\C, \Set]}) ) .$$ Even acknowledging that I am getting confused by the complicated "type signatures", I don't see any obvious candidates for $G_1, G_2$, and $\G$.

Possibly related questions:

Are fully faithful functors left-cancellable up to natural isomorphism?

Natural Transformation Between Covariant Hom-Functors

Adjoint Functors induce Natural Transformations

Natural transformation induced by adjoint functors.

Proving that the transformation obtained from an adjoint pair is natural

The Natural Isomorphism in Yoneda Lemma

Yoneda Lemma and isomorphisms

bivariate Yoneda lemma

Is the natural isomorphism in an adjunction uniquely determined by the pair of adjoint functors

Answer 2

The comments to the answer https://math.stackexchange.com/a/3922825/ indicate that the "diagonal Yoneda lemma" goes back to [MacLane, S. (1965). Categorical algebra. Bulletin of the American Mathematical Society, 71(1), 40-106] which can be found at https://www.ams.org/journals/bull/1965-71-01/S0002-9904-1965-11234-4/

In more modern terminology, the diagonal Yoneda lemma concerns extranatural transformations and representable profunctors. A (small) profunctor can be thought of as a functor $D(-,-)\colon\mathcal C^{op}\times\mathcal C\to\mathcal Set$.

Every (locally small) category $\mathcal C$ has an associated (small) representable profunctor $\mathcal C(-,-)\colon\mathcal C^{op}\times\mathcal C\to\mathcal Set$.

A family of elements $d_X\in D(X,X)$ is an extranatural transformation if $D(f,Y)(d_Y)=D(X,f)(d_X)$ for every morphism $f\colon X\to Y$.

Given functors $G_1,G_2\colon\mathcal C\to\mathcal G$, extranatural families $e_X\in\mathcal G(G_1-,G_2-)$ are exactly natural transformations $G_1\Rightarrow G_2$.

A more general version of your Lemma is that the image of an extrantural family under a natural transformation is extranatural.

This is indeed a corollary of the (diagonal) Yoneda lemma, whose statement is as follows.

Note that the identity natural transformation on the identity functor is given by the extranatural family of identity morphisms $\operatorname{id}_X\in\mathcal C(X,X)$. The image of the familiy $\operatorname{id}_X\in\mathcal C(X,X)$ under a natural transformation $\mathcal C(-,-)\Rightarrow\mathcal D(-,-)$ then determines an extranatural family $d_X\in D(X,X)$. This association of extranatural families of elements to natural transformations out of the representable profunctor is a bijection.

Since composition of natural transformations is a natural transformation, it follows that images of extranatural transformations under natural transformations are extranatural. This is the proof of the more general version of the Lemma from the diagonal Yoneda lemma.

The diagonal Yoneda lemma itself contains a special case of the more general lemma: that the image of the identity extranatural transformation is extranatural. The bijection aspect I don't think follows only from the diagonal Yoneda lemma.

Answer 3

This is to try to clarify / confirm the notation and definitions used in this answer. $\newcommand{\C}{\mathscr{C}}$$\newcommand{\F}{\mathscr{F}}$$\newcommand{\G}{\mathscr{G}}$$\newcommand{\D}{\mathscr{D}}$$\newcommand{\op}{\operatorname{op}}$$\newcommand{\Set}{\operatorname{Set}}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\Id}{\operatorname{Id}}$$\newcommand{\B}{\mathscr{B}}$$\newcommand{\Ob}{\operatorname{Ob}}$$\newcommand{\Mor}{\operatorname{Mor}}$

Definition: A profunctor $\mathbf{F}: \C \nrightarrow \D$ is a functor $\D^{\op} \times \C \to \Set$.

Definition: A profunctor $\mathbf{F}: \C \nrightarrow \D$ is called a representable profunctor if there exists a functor $F: \C \to \D$ such that $\mathbf{F} = \Hom_{\D} \circ ({\Id_{\D}}^{\op}, F) : \D^{\op} \times \C \to \Set$. We say that $\mathbf{F}$ represents $F$.

Definition: An endoprofunctor $\mathbf{F}$ is a profunctor whose domain and codomain category are the same, so $\mathbf{F}: \C \nrightarrow \C$, i.e. $\mathbf{F}: \C^{\op} \times \C \to \Set$ as a functor.

Fact: For any category $\C$, $\Hom_{\C}: \C \nrightarrow \C$ is a representable profunctor, representing $\Id_{\C}: C \to C$. It is of course also an endoprofunctor.

Fact: Given two functors $F_1, F_2: \C \to \D$ (i.e. two arbitrary functors with the same domain category and codomain category), the functor $\Hom_{\D} \circ (F_1^{\op}, F_2): \C^{\op} \times \C \to \Set$ is a profunctor $\C \nrightarrow \C$. (This is irrespective of the codomain category $\D$.)

Note: A lot of the following definitions can be generalized from (endo)profunctors, functors $\C^{\op} \times \C \to \Set$, to more general functors $\C^{\op} \times \C \to \D$ for an arbitrary category $\D$. But that level of generality is not needed here, so I will usually assume $\D = \Set$.

Definition: An extranatural transformation $\alpha: \mathbf{F} \nRightarrow \mathbf{G}$ from an endoprofunctor $\mathbf{F}: \B \nrightarrow \B$ to an endoprofunctor $\mathbf{G}: \C \nrightarrow \C$ consists of the following data:

• for every object $(b,c) \in \Ob(\B \times \C)$, a function $\alpha_{b,c} : \mathbf{F}(b,b) \to \mathbf{G}(c,c)$

satisfying the following two constraints:

• (cowedge condition) ("extranaturality in $b$") for every morphism $b_1 \overset{f}{\rightarrow} b_2 \in \Mor(\B)$ and every $c \in \Ob(\C)$ the following square in $\Set$ commutes: $$ \require{AMScd} \begin{CD} \mathbf{F}(b_2, b_1) @> \displaystyle \mathbf{F}(f^{\op}, \Id_{b_1}) >> \mathbf{F}(b_1, b_1) \\ @V \displaystyle \mathbf{F}(\Id_{b_2}^{\op}, f) VV @VV \displaystyle \alpha_{b_1, c} V \\ \mathbf{F}(b_2, b_2) @> \displaystyle \alpha_{b_2, c} >> \mathbf{G}(c,c) \end{CD} $$ writing $b_1 \overset{f^{\op}}{\leftarrow} b_2$ for the corresponding morphism $\in \Mor(\B^{\op})$.

• (wedge condition) ("extranaturality in $c$") for every $b \in \Ob(\B)$ and every morphism $c_1 \overset{g}{\rightarrow} c_2 \in \Mor(\C)$ the following square in $\Set$ commutes: $$ \require{AMScd} \begin{CD} \mathbf{F}(b, b) @> \displaystyle \alpha_{b,c_1} >> \mathbf{G}(c_1, c_1) \\ @V \displaystyle \alpha_{b, c_2} VV @VV \displaystyle \mathbf{G}(\Id_{c_1}^{\op}, g) V \\ \mathbf{G}(c_2, c_2) @> \displaystyle \mathbf{G}(g^{\op}, \Id_{c_2}) >> \mathbf{G}(c_1,c_2) \end{CD} $$ writing $c_1 \overset{g^{\op}}{\leftarrow} c_2$ for the corresponding morphism $\in \Mor(\C^{\op})$.

Definition: An extranatural family (called "diagonal spread" in MacLane's paper, cf. bottom of p. 54 in section 7) for an endoprofunctor $\mathbf{F}: \C \nrightarrow \C$ consists of the following data:

• for every $c \in \Ob(\C)$, an element $\alpha_c \in \mathbf{F}(c,c)$

satisfying the constraint:

• for every morphism $c_1 \overset{f}{\rightarrow} c_2 \in \Mor(\C)$, one has that ${\mathbf{F}(\Id_{c_1}^{\op}, f)(\alpha_{c_1}) = \mathbf{F}(f^{\op}, \Id_{c_2}) (\alpha_{c_2})}$.

Fact: Let $\ast$ denote the terminal category. (By abuse of notation, $\ast$ will also denote the unique object of the terminal category.) Then an extranatural family for an endoprofunctor $\mathbf{F}: \C \nrightarrow \C$ is equivalent to an extranatural transformation $\mathbf{Id}_{\ast} \nRightarrow \mathbf{F}$.

Here $\mathbf{Id}_{\ast}$ denotes the representable profunctor (as a functor $\Hom_{\ast}:\ast^{\op} \times \ast \to \Set$) that represents the identity functor $\Id_{\ast}: \ast \to \ast$. In particular, note that there is an obvious bijection between $\Ob(\ast \times \C)$ and $\Ob(\C)$, and that $\Hom_{\ast}(\ast, \ast) = \{\Id_{\ast}\}$. (This is another abuse of notation, where the identity morphism for the object $\ast$ is denoted the same as the identity functor of the category $\ast$.) A singleton set like $\{\Id_{\ast}\}$ is a terminal object in $\Set$, and morphisms from a terminal object in $\Set$ to an arbitrary set $X$ are equivalent to choosing an element of the set $X$ (due to the well-pointedness property of $\Set$). So we get two commutative diagrams:

$$ \require{AMScd} \begin{CD} \Hom_{\ast}(\ast, \ast) @> \displaystyle \Hom_{\ast}(\Id_{\ast}^{\op}, \Id_{\ast}) >> \Hom_{\ast}(\ast, \ast) \\ @V \displaystyle \Hom_{\ast}(\Id_{\ast}^{\op}, \Id_{\ast}) VV @VV \displaystyle \alpha_{\ast, c} V \\ \Hom_{\ast}(\ast, \ast) @> \displaystyle \alpha_{\ast, c} >> \mathbf{G}(c,c) \end{CD} $$ and $$ \require{AMScd} \begin{CD} \Hom_{\ast}(\ast, \ast) @> \displaystyle \alpha_{\ast,c_1} >> \mathbf{F}(c_1, c_1) \\ @V \displaystyle \alpha_{\ast, c_2} VV @VV \displaystyle \mathbf{F}(\Id_{c_1}^{\op}, f) V \\ \mathbf{F}(c_2, c_2) @> \displaystyle \mathbf{F}(f^{\op}, \Id_{c_2}) >> \mathbf{F}(c_1,c_2) \end{CD} $$ The first commutative diagram is trivial / degenerate, and just says that the function $\alpha_{\ast, c}$ in $\Set$ (which we can identify with the element $\alpha_c$ of $\mathbf{F}(c,c)$ that it evaluates to) is equal to itself, which of course it is. The second diagram says that ${\mathbf{F}(\Id_{c_1}^{\op}, f) \circ \alpha_{\ast, c_1} = \mathbf{F}(f^{\op}, \Id_{c_2}) \circ \alpha_{\ast, c_2}}$, which of course is equivalent to the condition defining an extranatural family for $\mathbf{F}$.

Fact: Given functors $G_1, G_2: \C \to \G$, an extranatural family for the functor $\Hom_{\G} \circ (G_1^{\op}, G_2) : \C^{\op} \times \C \to \Set$ (considered as an endoprofunctor $\C \nrightarrow \C$) is the same thing as a natural transformation $G_1 \Rightarrow G_2$.

(This fact is the comment immediately following the proof of theorem 7.5 in MacLane's paper.)

Basically look at the non-trivial commutative diagram from the proof that extranatural families are a special case of extranatural transformations, setting $\mathbf{F} = \Hom_{\G} \circ (G_1^{\op}, G_2)$: $$ \require{AMScd} \begin{CD} \Hom_{\ast}(\ast, \ast) @> \displaystyle \alpha_{\ast,c_1} >> \Hom_{\G}(G_1(c_1), G_2(c_1)) \\ @V \displaystyle \alpha_{\ast, c_2} VV @VV \displaystyle \Hom_{\G}(\Id_{G_1(c_1)}^{\op}, G_2(f)) V \\ \Hom_{\G}(G_1(c_2), G_2(c_2)) @> \displaystyle \Hom_{\G}(G_1(f)^{\op}, \Id_{G_2(c_2)}) >> \Hom_{\G}(G_1(c_1), G_2(c_2)) \end{CD} $$ Then the result follows by using the definition of the $\Hom$ functor on morphisms. Basically the above commutative diagram says pick (via $\alpha_{\ast,c_1}$) a morphism $\alpha_{c_1} \in \Hom_{\G}(G_1(c_1), G_2(c_1))$, i.e. $G_1(c_1) \overset{\alpha_{c_1}}{\rightarrow} G_2(c_1)$, likeiwse pick a morphism $G_1(c_2) \overset{\alpha_{c_2}}{\rightarrow} G_2(c_2)$, and then because the "action" of Hom functors on the "left coordinate" for morphisms is pre-composing, and for the "right coordinate" is post-composing, we get from the above diagram that: $$\begin{array}{rcl} G_2(f) \circ \alpha_{c_1} &= &G_2(f) \circ \alpha_{c_1} \circ \Id_{c_1} \\ &= & \Hom_{\G}(\Id_{G_1(c_1)}^{\op}, G_2(f)) (\alpha_{c_1}) \\ &=& \Hom_{\G}(G_1(f)^{\op}, \Id_{G_2(c_2)}) (\alpha_{c_2}) \\ & = & \Id_{G_2(c_2)} \circ \alpha_{c_2} \circ G_1(f) \\ & = & \alpha_{c_2} \circ G_1(f) \end{array}$$ and of course the equation $G_2(f) \circ \alpha_{c_1} =\alpha_{c_2} \circ G_1(f)$ is the same thing as the naturality square of a natural transformation:

$$ \require{AMScd} \begin{CD} G_1(c_1) @> \displaystyle G_1(f) >> G_1(c_2) \\ @V \displaystyle \alpha_{c_1} VV @VV \displaystyle \alpha_{c_2} V \\ G_2(c_1) @> \displaystyle G_2(f) >> G_2(c_2) \end{CD} $$

Fact: Given endoprofunctors $\mathbf{F}: \B \nrightarrow \B$, $\mathbf{G}: \C \nrightarrow \C$, and $\mathbf{H}: \C \nrightarrow \C$ (notice that the category associated to $\mathbf{G}$ and $\mathbf{H}$ must be the same, but not for $\mathbf{F}$), one can define the "composition" $\beta \circ \alpha$ of an extranatural transformation $\alpha: \mathbf{F} \nRightarrow \mathbf{G}$ with a natural transformation $\beta: \mathbf{G} \Rightarrow \mathbf{H}$ (considering $\mathbf{G}$, $\mathbf{H}$ as functors) that is an extranatural transformation $\beta \circ \alpha: \mathbf{F} \nRightarrow \mathbf{H}$.

Aside: Notice that the definition below is actually fairly "weird" to the extent that it ignores / does not use the vast majority of the structure of the natural transformation $\beta$, instead only using the "diagonal" components $\beta_{c,c}$ and not using / imposing any requirements on any other part of the structure of $\beta$. (This seems to be reflected in the notation used in Exercise 1.4 here.) Of course perhaps that should be expected to the extent that the notion of "extranatural transformation" (or at least each of the cowedge condition and wedge condition considered separately) does not use much / all of the structure of the endoprofunctor either. So quite probably these definitions apply to more general constructions, although it's questionable whether the extra generality would be useful.

Anyway, for the "cowedge condition" / "extranaturality in $b$, the proof of this is easy. We start with

$$ \require{AMScd} \begin{CD} \mathbf{F}(b_2, b_1) @> \displaystyle \mathbf{F}(f^{\op}, \Id_{b_1}) >> \mathbf{F}(b_1, b_1) \\ @V \displaystyle \mathbf{F}(\Id_{b_2}^{\op}, f) VV @VV \displaystyle \alpha_{b_1, c} V \\ \mathbf{F}(b_2, b_2) @> \displaystyle \alpha_{b_2, c} >> \mathbf{G}(c,c) \end{CD} $$

which is equivalent to the equation $\alpha_{b_2, c} \circ \mathbf{F}(\Id_{b_2}^{\op}, f) = \alpha_{b_1, c} \circ \mathbf{F}(f^{\op}, \Id_{b_1})$, which of course implies that $\beta_{c,c,} \circ \alpha_{b_2, c} \circ \mathbf{F}(\Id_{b_2}^{\op}, f) = \beta_{c,c} \circ \alpha_{b_1, c} \circ \mathbf{F}(f^{\op}, \Id_{b_1})$, leading to the cowedge condition being satisfied for $\beta \circ \alpha$:

$$ \require{AMScd} \begin{CD} \mathbf{F}(b_2, b_1) @> \displaystyle \mathbf{F}(f^{\op}, \Id_{b_1}) >> \mathbf{F}(b_1, b_1) \\ @V \displaystyle \mathbf{F}(\Id_{b_2}^{\op}, f) VV @VV \displaystyle \beta_{c,c} \circ \alpha_{b_1, c} V \\ \mathbf{F}(b_2, b_2) @> \displaystyle \beta_{c, c} \circ \alpha_{b_2, c} >> \mathbf{H}(c,c) \end{CD} $$

The wedge condition is more subtle (and the only place where we actually need the naturality of $\beta$ and non-diagonal components). Basically we want to start with the wedge condition / extranaturality in $c$ square:

$$ \require{AMScd} \begin{CD} \mathbf{F}(b, b) @> \displaystyle \alpha_{b,c_1} >> \mathbf{G}(c_1, c_1) \\ @V \displaystyle \alpha_{b, c_2} VV @VV \displaystyle \mathbf{G}(\Id_{c_1}^{\op}, g) V \\ \mathbf{G}(c_2, c_2) @> \displaystyle \mathbf{G}(g^{\op}, \Id_{c_2}) >> \mathbf{G}(c_1,c_2) \end{CD} $$

and adjoin to it the two naturality squares:

$$ \require{AMScd} \begin{CD} \mathbf{G}(c_2, c_2) @> \displaystyle \mathbf{G}(g^{\op}, \Id_{c_2}) >> \mathbf{G}(c_1,c_2) \\ @V \displaystyle \beta_{c_2, c_2} VV @VV \displaystyle \beta_{c_1, c_2} V \\ \mathbf{H}(c_2, c_2) @> \displaystyle \mathbf{H}(g^{\op}, \Id_{c_2}) >> \mathbf{H}(c_1,c_2) \end{CD} $$

and

$$ \require{AMScd} \begin{CD} \mathbf{G}(c_1, c_1) @> \displaystyle \mathbf{G}(\Id_{c_1}^{\op}, g) >> \mathbf{G}(c_1,c_2) \\ @V \displaystyle \beta_{c_1, c_1} VV @VV \displaystyle \beta_{c_1, c_2} V \\ \mathbf{H}(c_1, c_1) @> \displaystyle \mathbf{H}(\Id_{c_1}^{\op}, g) >> \mathbf{H}(c_1,c_2) \end{CD} $$

while "gluing" along the morphism $\mathbf{G}(c_1,c_2) \overset{\beta_{c_1, c_2}}{\rightarrow} \mathbf{H}(c_1,c_2)$. (Note that you need to "flip" the second commutative square across its upper-left-to-bottom-right diagonal in order to do this.) The result is a "hexagon" or, if you allow popping out into the 3rd dimension, half of a cube.

(Notice that the diagram in Remark 1.1.7 of "This is the co/end" appears to be the limit of such a diagram, so two "half-cubes" connected together to form a while cube -- which makes sense because it depicts a "universal wedge" i.e. an "end".)

If you ignore the "inner" three morphisms (that all have $\mathbf{G}(c_1, c_2)$ as either domain or codomain) you get a new extranaturality square, i.e. one satisfying the required wedge condition / extranaturality in $c$ for $\mathbf{F} \nRightarrow \mathbf{H}$:

$$ \require{AMScd} \begin{CD} \mathbf{F}(b, b) @> \displaystyle \beta_{c_1, c_1} \circ \alpha_{b,c_1} >> \mathbf{H}(c_1, c_1) \\ @V \displaystyle \beta_{c_2, c_2} \circ \alpha_{b, c_2} VV @VV \displaystyle \mathbf{H}(\Id_{c_1}^{\op}, g) V \\ \mathbf{H}(c_2, c_2) @> \displaystyle \mathbf{H}(g^{\op}, \Id_{c_2}) >> \mathbf{H}(c_1,c_2) \end{CD} $$

Hence defining $\beta \circ \alpha$ as the above, we have shown that $\beta \circ \alpha$ is an extranatural transformation (extranatural in both $b$ and $c$) $\mathbf{F} \nRightarrow \mathbf{H}$).

Corollary: The lemma that this question is about is corollary of the above facts. Specifically, given functors $G_1, G_2: \C \to \G$, a natural transformation $G_1 \Rightarrow G_2$ is the same thing as an extranatural family for $\Hom_{\G} \circ (G_1^{\op}, G_2)$, considering that functor $\C^{\op} \times \C \to \Set$ as an endoprofunctor $\mathbf{G} :\C \nrightarrow \C$, which in turn is the same thing as an extranatural transformation $\alpha: \mathbf{Id}_{\ast} \nRightarrow \mathbf{G}$. Then given functors $F_1, F_2: \C \to \F$ and a natural transformation $\beta: Hom_{\G} \circ (G_1^{\op}, G_2) \Rightarrow Hom_{\F} \circ (F_1^{\op}, F_2)$, we get an extranatural transformation $\beta \circ \alpha: \mathbf{Id}_{\ast} \nRightarrow \mathbf{F}$, where $\mathbf{F}: \C \nrightarrow \C$ is the endoprofunctor notation for the functor $\Hom_{\F} \circ (F_1^{\op}, F_2): \C^{\op} \times \C \to \Set$. The extranatural transformation $\beta \circ \alpha: \mathbf{Id}_{\ast} \nRightarrow \mathbf{F}$ is the same as an extranatural family for $\mathbf{F}$, which because $\mathbf{F} = \Hom_{\F} \circ (F_1^{\op}, F_2)$ is the same thing as a natural transformation $F_1 \Rightarrow F_2$.