How to understand the Yoneda lemma

Question

I have difficulty understanding some points of the proposition,how can the natural transformation τ entirely determined by the element of F（A）by proving that τB（φ）=（Fφ）（τA）（1A）？And if the equation is held，how can it sets up a bijection？Thank you for your help！

Here is the proof

Answer 1

The diagram in the proof commutes because $\tau$ is assumed to be a natural transformation $\mathcal C(A, -) \to F$.
This, applied to $1_A\in\mathcal C(A, A)$ yields just the given equation $$\tau_B(\varphi) =F(\varphi)(\tau_A(1_A)) $$ (using $\mathcal C(A,\varphi) =\psi\mapsto \varphi\circ\psi$),
which shows that any value of $\tau$ can indeed be obtained by using $F$ and $\tau_A(1_A)$.

It already implies that the assignment $\tau\mapsto \tau_A(1_A)$ is injective.

To prove it is surjective, they construct a $\tau$ for a given arbitrary element $x\in F(A)$ such that $\tau_A(1_A)=x$.
[Specifically, by the above, it has to be of the form $\tau_B(\varphi) := F(\varphi)(x)$, and we can check it indeed defines a natural transformation.]

Answer 2

how can the natural transformation $\tau$ entirely determined by the element of $F(A)$ by proving that $\tau_B(\phi) = (F\phi)(\tau_A)(1_A)$?

The natural transformation $\tau$ is "entirely determined" if for every $B \in \mathfrak C$, we know what the morphism $\tau_B$ is.

Note that the target category of our two functors is $\mathfrak S$, the category of sets. So, $\tau_B$ should be a function from $\mathfrak C(A,B)$ to $F(B)$. If we know what $\tau_A(1_A)$ is, the $\tau_B$ is determined, which is to say that we can figure out the output $\tau_B(\phi)$ for any input $\phi$. As the diagram implies, for $\phi \in \mathfrak C(A,B)$, we have $$ \tau_B(\phi) = (\tau_B \circ \phi_*)(1_A) = (F\phi \circ \tau_A)(1_A) = (F\phi)(\tau_A(1_A)). $$

how can it set up a bijection?

We want to show that the map $\tau \mapsto \tau_A(1_A)$ is bijective. What we just showed above is that if $\sigma,\tau$ are two natural transformations with $\tau_A(1_A) = \sigma_A(1_A)$, then we must have $\tau_B = \sigma_B$ for all objects $B$ of $\mathfrak C$. In other words, we must have $\tau = \sigma$. So, we just showed that the map is injective.

To show that the map is surjective, we need to show that for any $x \in F(A)$, there exists a natural transformation such that $\tau_A(1_A) = x$. In order to do that, it suffices to note that the functions defined by $\tau_B(\phi) = (F\phi)(x)$ so that we have $$ \tau_B(\phi) = (F\phi \circ \tau_A)(1_A) = (F\phi)(\tau_A(1_A)) = (F\phi)(x) $$ specify a natural transformation $\tau$ for which $\tau_A(1_A) = x$.