I've been reading through Rotman's Homological Algebra, and part i of corollary 1.18 states the following: $\DeclareMathOperator{\hom}{Hom}$

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

The part I'm having trouble with is understanding why the last statement is true. The proof in the book merely states, "The uniqueness assertion follows from injectivity of the Yoneda function $y$."

In trying to prove uniqueness, I came up with the following:

Suppose $\tau_C=\theta^*$. Then that means $\psi^*=\theta^*$. So for every $f:A\rightarrow C$, $f\psi=f\theta$. How can we conclude from here that $\psi=\theta$?

  • $\begingroup$ My understanding is that this has to be shown for each $C$ separately. $\endgroup$ – Carmeister Oct 16 '16 at 19:16

The map $\psi$ does not depend on $C$; so you can use $C=A$ and $f=1_A$.

The notation is somewhat misleading; for every $C$ you have a different map $\psi^*\colon\operatorname{Hom}_{\mathcal{C}}(A,C)\to\operatorname{Hom}_{\mathcal{C}}(B,C)$, but $\psi$ itself is fixed.

  • $\begingroup$ Ah, I think understand now. So the full statement of the uniqueness criterion is that if there exists a $\theta$ such that $\tau_C=\theta^*$ for all $C$, then $\theta=\phi$, yes? $\endgroup$ – Carmeister Oct 16 '16 at 20:15
  • $\begingroup$ @Carmeister Yes, the same $\theta$ is supposed to work for all $C$. $\endgroup$ – egreg Oct 16 '16 at 20:23

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