Why $\Psi\circ\Phi=1$ in Yoneda lemma?

Question

I am trying to understand the proof: How can one show that $\Psi\circ\Phi=1_{\textrm{Nat}(A(A,-),F)}$?

Let $\tau:A(A,-)\rightarrow F$ be a natural transformation. Then $(\Psi\circ\Phi)(\tau)=\Psi(\tau_A(1_A))=\{\Psi(\tau_1(1_A))_x\}_{X\in\textrm{Ob} X}$

And here I'm stuck. How to show that it equals $\tau$?

P.S.: I've just started studying category theory, so multi-layer constructions seem really complex to me.

Answer 1

In order to show that $(\Psi \circ \Phi)(\tau) = \tau$, we must show that for every $X \in \text{Ob}(\mathcal A)$, $\tau_X : \mathcal A(A,X)\to F(X)$ (the component of $\tau$ at $X$) is equal to $(\Psi \circ \Phi)(\tau)_X$. Note that both $\tau_X$ and $(\Psi \circ \Phi)(\tau)_X$ are morphisms in $\mathsf{Sets}$ from the set $\mathcal A(A,X)$ to the set $F(X)$. So, to show that these two set maps are equal, we want to show that for every $f$ in the domain (that is, every $f:A \to X$), we have $$ \tau_X(f) = (\Psi\circ \Phi)(\tau)_X(f). $$ With that goal, note that $$ \begin{align} (\Psi\circ \Phi)(\tau)_X(f) &= [\{\Psi(\tau_A(1_A))_{X'}\}_{X' \in \text{Ob}(\mathcal A)}]_X(f) \\ & = \Psi(\tau_A(1_A))_X(f) \\ & = F(f)(\tau_A(1_A)) \\ & = (F(f) \circ \tau_A)(1_A) \\ & = (\tau_X \circ [\mathcal A(A,-)(f)])(1_A) \\ & = \tau_X([\mathcal A(A,-)(f)](1_A)) = \tau_X(f \circ 1_A) = \tau_X(f). \end{align} $$