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 $\theta = \psi.$


(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?


1 Answer 1


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 ^_^

  • $\begingroup$ 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. $\endgroup$
    – IsaacR24
    Sep 17 at 11:44
  • $\begingroup$ 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? $\endgroup$
    – IsaacR24
    Sep 17 at 11:47

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