# Corollary to Yoneda Lemma

I'm working from Rotman's Homological Algebra text. There's just one detail of the Corollary that I don't follow.

## 1. Yoneda Lemma

Theorem 1.17 (Yoneda Lemma). Let $$\mathcal{C}$$ be a category, let $$A \in \text{obj}(\mathcal{C})$$, and let $$G : \mathcal{C} \to \mathsf{Sets}$$ be a covariant functor. Then there is a bijection

$$y : \text{Nat}(\text{Hom}_\mathcal{C}(A,\square), G) \to G(A)$$

given by $$y : \tau \mapsto \tau_A(1_A).$$

## 2. Corollary

Corollary 1.18. Let $$\mathcal{C}$$ be a category and let $$A, B \in \text{obj}(\mathcal{C})$$.

(i) If $$\tau : \text{Hom}_\mathcal{C}(A,\square) \to \text{Hom}_\mathcal{C}(B,\square)$$ is a natural transformation, then for all $$C \in \text{obj}(\mathcal{C})$$, we have $$\tau_C = \psi^*$$ where $$\psi = \tau_A(1_A) : B \to A$$ and $$\psi^*$$ is the induced map $$\text{Hom}_\mathcal{C}(A,C) \to \text{Hom}_\mathcal{C}(B,C)$$ given by $$\varphi \mapsto \varphi \psi$$. Moreover, the morphism $$\psi$$ is unique: if $$\tau_C = \theta^*$$, then $$\theta = \psi.$$

Proof.

(i) If we denote $$\tau_A(1_A) \in \text{Hom}_\mathcal{C}(B,A)$$ by $$\psi$$, then the Yoneda Lemma says, for all $$C \in \text{obj}(\mathcal{C})$$ and all $$\varphi \in \text{Hom}_\mathcal{C}(A,C)$$, that $$\tau_C(\varphi) = \varphi_*(\psi)$$. But $$\varphi_*(\psi) = \varphi \psi = \psi^*(\varphi)$$. The uniqueness assertion follows from injectivity of the Yoneda function $$y$$

## 3. Question

I don't see why the uniqueness of $$\psi$$ follows from the injectivity of $$y$$. We have $$\psi^* = \tau_C = \theta^*$$. In other words, for all $$\phi \in \text{Hom}(A,C)$$, $$\phi \psi = \phi \theta$$.

We can also say that $$\theta^* = y(\tau)^*$$. If we knew that $$\text{Hom}(A,C)$$ contained an injective morphism, the result would certainly follow. What am I missing?

The yoneda embedding $$y$$ eats an object $$C$$ of $$\mathcal{C}$$ and spits out a functor $$y(C) = \text{Hom}_\mathcal{C}(C, \square) : \mathcal{C} \to \mathsf{Set}$$. Now the yoneda lemma tells us that for any other functor $$F : \mathcal{C} \to \mathsf{Sets}$$ we have a bijection

$$\text{Nat}(y(C), F) \cong F(C)$$

(remember that $$\text{Nat}(y(C),F)$$ is the set of natural transformations, and $$F(C)$$ is the set $$F$$ assigns to the object $$C$$).

Now let's look at $$\text{Nat}(\text{Hom}_\mathcal{C}(A,\square), \text{Hom}_\mathcal{C}(B,\square))$$, which we can write as $$\text{Nat}(y(A), \text{Hom}_\mathcal{C}(B,\square))$$. The yoneda lemma says that this set of natural transformations is in bijection with $$\text{Hom}_\mathcal{C}(B,\square)(A) = \text{Hom}_\mathcal{C}(B,A)$$. In fact, it tells us the bijection!

To each $$\tau \in \text{Nat}(\text{Hom}_\mathcal{C}(A,\square), \text{Hom}_\mathcal{C}(B,\square))$$, we assign the arrow $$\psi = \tau_A(1_A) \in \text{Hom}_\mathcal{C}(B,A)$$.

Conversely, to each $$\psi \in \text{Hom}_\mathcal{C}(B,A)$$ we assign the natural transformation $$\tau = \psi^* \in \text{Nat}(\text{Hom}_\mathcal{C}(A,\square), \text{Hom}_\mathcal{C}(B,\square))$$.

Let's mainly pay attention to the map $$\psi \mapsto \psi^*$$. What does it mean that this map is a bijection? Surjectivity tells us that every natural transformation $$\tau$$ is $$\psi^*$$ for some $$\psi$$. Injectivity says exactly that the $$\psi$$ associated with some $$\tau$$ is unique! Indeed, if $$\psi_1$$ and $$\psi_2$$ are different arrows in $$\text{Hom}_\mathcal{C}(B,A)$$, they must get sent to different natural transformations $$\tau_1 = \psi_1^*$$ and $$\tau_2 = \psi_2^*$$.

I hope this helps ^_^

• Thanks, I think I understand but just 2 clarifications. I believe you're referring to the Yoneda embedding $Y$ which is a functor. I'm just about to review this corollary but haven't yet. Sep 17 at 11:44
• Second, we don't have $\tau = \psi^*$ but instead $\tau_C = \psi^*$. But then $\psi \mapsto \tau \mapsto \tau_C$ is a composition of injective functions, so that $\psi$ must be unique. $\tau \mapsto \tau_C$ must be injective, otherwise the hom sets would not be disjoint (a requirement of categories). Did I get this all right? Sep 17 at 11:47