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Can anyone explain to me Yoneda Lemma proof in great details? i.e. they usually say " ... it is easy to see that these morphisms are inverse to each other.." without explanation.

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    Perhaps it would be better if you explained what parts of the Yoneda lemma's proof you understand, and we can help with the bits that are unclear? If the only problem is understanding why the Yoneda embedding is fully faithful, there are two steps. For faithful, if a morphism $f$ is sent to a natural transformation $\eta$ in the functor category, we can recover $f$ by applying $\eta$ to the identity map. Therefore, no two different maps can be sent to the same transformation. For full, the definition of natural transformation implies that the image of the identity map determines everything.2011-12-25
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    You would learn more if you try it yourself.2012-01-05
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    Probably the most detailled proof can be found in Tom Leinster's new book "Basic Category theory".2015-04-23

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A presheaf $F$ is simply a contravarient functor $C^{op} \to \text{Sets}$.

Please refer to these two sources for more details on the proof. These are the two that, combined, I found most useful:

Now I will reconstruct from my notes, what I've learned. Let $C$ be a locally small category.

Let $C(-, c) \equiv \text{Hom}_C(-, c) \equiv \text{Mor}_C(-, c) \equiv h_c$. These are all just notations for the same thing, the contravariant hom functor. $C(-, c)$ is contravariant and $C(c, -)$ is covariant. Make sure you understand those first.

For $f : y \to x$ in $C$ we have the morphism of hom sets $C(x, c) \xrightarrow{C(f,c)} C(y,c)$ where $C(f, c) \equiv - \circ f$ ie composition by $f$ on the right. Clearly composition on the right would change the domain, so to remember $C(f, c) \equiv -\circ f$ just remember that $f$ is in the place of the domain in the expression $C(f, c)$.

$C(-,c)$ is a presheaf. The presheaves form the category $\text{Psh}(C) = \text{Fun}(C^{op}, \text{Sets})$ where morphisms are natural transformations of functors. By defintion a natural transformation of functors $F \xrightarrow{\phi} G$, where $F, G : C \to D$ is a collection of maps $\{\phi_x\}_{x \in \text{Ob}(C)}$ such that the following commutes for every $f : y \to x$ in $C$:

$$ \begin{matrix} F(y) & \xrightarrow{F(f)} & F(x) \\ \downarrow \phi_y & \ & \downarrow\phi_x\\ G(y) & \xrightarrow{G(f)} & G(x) \end{matrix} $$

Now, for any morphism $\phi : u \to v $ in $C$ there corresponds a natural map of hom functors $h_u \to h_v$ (now we've switched notation; see above) given by composing with $\phi$: i.e. by letting each $\phi_z = \phi$. We will show this.

$$ g \in h_u(y) = \text{Mor}_C(y, u) \\ \implies h_u(f)\circ \phi_y (g) = h_u(f) \circ (\phi\circ g) = \phi\circ g \circ f $$ Remember that $h_u(f) \equiv -\circ f$. Similarly, $$ \phi_x \circ h_v(f) (g) = \phi_x \circ (g\circ f) = \phi\circ g\circ f $$

So that we have a similar naturality diagram to the above: $$ \begin{matrix} h_u(y) & \xrightarrow{h_u(f)} & h_u(x) \\ \downarrow \phi_y & \ & \downarrow \phi_x \\ h_v(y) & \xrightarrow{h_v(f)} & h_v(x) \end{matrix} $$

Such an induced natural map is given the notation $h(\phi)$. So that resolves that confusion I personally had. In other words $h(\phi)_z \equiv \phi_z = \phi$.


Now the first part of the Yoneda lemma from the Stacks Project is:

Lemma 4.3.5 (Yoneda lemma). Let $u,v \in \text{Ob}(C)$. Given any morphism of functors $s : h_u \to h_v$ there is a unique morphism $\phi : u \to v$ such that $h(\phi) = s$.

Proof. In the previous diagram above, replace $\phi$ with $s$ to match up notations: $$ \begin{matrix} h_u(y) & \xrightarrow{h_u(f)} & h_u(x) \\ \downarrow s_y & \ & \downarrow s_x \\ h_v(y) & \xrightarrow{h_v(f)} & h_v(x) \end{matrix} $$

The above diagram must be satisfied for all $f : y \to x$ for $s$ to be a "morphism of functors" (a natural transformation of functors). Knowing that, the try the diagram out on $x = u$ itself! In other words let $f : x \to u$ be any arrow in $C$ with codomain $u$:

$$ \begin{matrix} h_u(u) & \xrightarrow{h_u(f)} & h_u(x) \\ \downarrow s_u & \ & \downarrow s_x \\ h_v(u) & \xrightarrow{h_v(f)} & h_v(x) \end{matrix} $$

Now take (define) $\phi = s_u(1_u)$. Then by this diagram above we already have $h_v(f)\circ s_u = s_v\circ h_u(f)$ so that we have $\phi_v \circ f = h_v(f) \circ \phi_v = h_v(f) \circ s_u (1_u) = s_v \circ h_u(f) (1_u) = s_v \circ f$ (Remember $\phi_z = \phi \ \forall z$ including $z = v$). Whenever two maps agree on all maps into them, they must be equal maps. Thus $\phi_v = s_v$ or in other words the natural map $h(\phi)$ as defined above equals the natural map $s$, or $h(\phi) = s$. Now to show uniqueness, we only need to look at the fact that $h_v(f)\circ s_u(1_u) = s_u(f) = \phi_u(f)$. Clearly this must be the case for any $h(\phi') = s$ as well. Thus $\phi = s_u(1_u)$ is uniquely defined and such $\phi$ that $h(\phi) = s$ is then unique. $\square$.

That takes care of the first part.


Next,

In other words the functor $h$ is fully faithful.

The functor $h$ is as we have been using it: $$ \text{Ob}(C) \ni u \mapsto h_u \in \text{PSh}(C) \\ h : (\phi : u \to v) \mapsto (h(\phi) : h_u \to h_v) $$

Since for each $s$ in $\text{Psh}(C)$ we have that $h(\phi) = s$ clearly, the map is bijective. But ah! We've made a mistake in our reasoning, we only showed bijectivity onto the subcategory of presheaves of the form $h_u$ ie. the hom-sets. Well, it turns out that the exact same arguments as above can be restated for the commuting square:

$$ \begin{matrix} h_u(u) & \xrightarrow{h_u(f)} & F(x) \\ \downarrow s_u & \ & \downarrow s_x \\ h_v(u) & \xrightarrow{F(f)} & F(x) \end{matrix} $$

where $F$ is any presheaf. So that patches up that.


Now for the last part:

More generally, given any contravariant functor $F$ and any object $u$ of $C$, we have a natural bijection $$ \Phi:\text{Mor}_{\text{Psh}}(C)(h_u, F) \to F(u), \ s \mapsto s_u(1_u) $$

Note that one direction of the bijection $\Phi$ is already listed. We need to show that $\Phi$ is a bijection so it suffices to show an inverse map $\Psi$. Well, let $\xi \in F(u)$, then $\Psi(\xi) = s$ where $s_v : h_u(v) \to F(v) : \ (f: v\to u ) \to F(f)(\xi)$ are the components of the natural map $s$ they say in the Stacks Project, will work. Well (this is a little iffy, but:)

$$ (\Psi \circ \Phi(s))_v(f) = (\Psi \circ (s_u \circ 1_u))_v(f) = (F(f)(s_u \circ 1_u))_v = s_v \circ h_u(f) \circ 1_u = s_v(f) \\ \square $$

Thus $\Psi\circ\Phi(s) = s$ and we're done. I don't know what they mean by a natural bijection though. Thus we've proved the last part of Yoneda's Lemma sometimes expressed as: $$ \text{Nat}(h_u, F) \simeq F(u) $$