Understanding the proof of Yoneda Embedding Definitions
Recently I was going through the proof Yoneda Embedding from Steven Roman's An Introduction to the Language of Category Theory. But before I go into the problem that I am having with the proof let me first state Roman's definition of an embedding (I do so because according to this there is "no satisfactory and generally accepted definition of embeddings that is applicable in all categories"). 

Let $\mathcal{C},\mathcal{D}$ be two categories and $F:\mathcal{C}\to\mathcal{D}$ be a functor. Then,
Definition 1. The restriction of $F$ to $\text{hom}_\mathcal{C}(A,B)$ as a local arrow part of $F$ and by the object part of $F$ we mean $F:\text{Obj}(\mathcal{C})\to\text{Obj}(\mathcal{D})$ where $\text{Obj}(\mathcal{C})$ and $\text{Obj}(\mathcal{D})$ denotes the set of objects of respectively $\mathcal{C}$ and $\mathcal{D}$. 
Definition 2. $F$ is full if all of its local arrow parts are surjective.
Definition 3. $F$ is faithful if all of its local arrow parts are injective.
Definition 4. $F$ is fully faithful if all of its local arrow parts are bijective.
Definition 5. $F$ is an embedding of $\mathcal{C}$ in $\mathcal{D}$ if it is fully faithful and the object part of $F$ is injective.

Background
Now let me come directly to my problem. What I was going through is the following theorem (number unaltered),

Theorem 48
Let $\mathcal{C}$ be a category. The contravariant functor $y: \mathcal{C} \to \sf{Set}^\mathcal{C}$ defined by,
  $$y(A)=\text{hom}_\mathcal{C}(A,\cdot)$$and, $$y(h)=\{{h_X}^\rightarrow\}:\text{hom}_\mathcal{C}(A,\cdot)\to\text{hom}_\mathcal{C}(B,\cdot)$$for all $A\in\mathcal{C}$ and all $h\in\text{hom}_\mathcal{C}(B,A)$
  is a contravariant embedding of $\mathcal{C}$ into the functor
  category $\sf{Set}^\mathcal{C}$, called the Yoneda embedding of $\mathcal{C}$ in $\mathsf{Set}^\mathcal{C}$.

Where ${h_X}^\rightarrow:\text{hom}_\mathcal{C}(A,X)\to\text{hom}_\mathcal{C}(B,X)$ defined by, $${h_X}^\rightarrow(g)=g\circ h_X$$
In the proof he writes,


*

*The object part of $y$ maps $A$ to $\text{hom}_\mathcal{C}(A,\cdot)$ and since $\text{hom}_\mathcal{C}(A,\cdot)$ and $\text{hom}_\mathcal{C}(B,\cdot)$ are distinct for distinct objects $A$ and $B$, the object part of $y$ is injective.

*As to injectivity, if $y_{A,B}(h) = y_{A,B}(k)$ for $h,k: B \to A$, then
$$\{h_X^\to:X\in\mathcal{C}\}=\{k_X^\to:X\in\mathcal{C}\}$$
In particular, for the components associated with $X=A$, we can apply them to $1_A$ (the identity morphism from $A$ to $A$) to get
$$1_A ∘ h = 1_A ∘ k$$
and so $h = k$.
My Questions


*

*How does it follow that $A\ne B$ implies that $\text{hom}_\mathcal{C}(A,\cdot)\ne\text{hom}_\mathcal{C}(B,\cdot)$?

*If $\{h_X^\to:X\in\mathcal{C}\}=\{k_X^\to:X\in\mathcal{C}\}$ then all we can say is the following, $$1_A ∘ h_Y = 1_A ∘ k_Z$$for some $Y,Z\in\mathcal{C}$. Why can we remove the subscripts? 
 A: Yes - you understood my comment on 1. Point 2, is similar. Namely: Suppose we are given $h\colon B \to A$. Then $h$ induces, for each $X$, a  function $$h_X\colon {\rm hom} (A, X) \to {\rm hom }( B, X), $$ defined by the rule $ h_X (g) = g \circ h $. [No subscript on the right.] 
[ Continuing the proof of Yoneda - although you're not asking for it: ]
Suppose we  are given 'another' morphism $k\colon B \to A$ (same source and target as $h$). 
That means that we  now have, for  each $X$, two functions
 $$h_X,k_X \colon {\rm hom} (A, X) \to {\rm hom }( B, X). $$  
But what  happens, if for each $X$, $h_X$ and $k_X$ coincide? That is, what happens if, for each $X$,  $$h_X = k_X?$$ 
Then the  previous equality holds, in particular, with $X=A$. That is,  $$h_A = k_A.$$ 
That means, for every $g\in {\rm hom} (A, A)$, that 
$$ h_A (g) = k_A(g),$$ or equivalently $g \circ h = g \circ k$. In particular, we can take $g= 1_A$. Therefore $1_A\circ h = 1_A\circ k$, and $h = k$ - and we are done.
To summarize: if $h_X=k_X$ for every $X$, then $h_A=k_A$, so that $h_A(1_A)=k_A(1_A)$, so that $1_A\circ h = 1_A\circ k$, so that $h=k$.
[This answer was a follow-up to a comment above - but was itself too long for a comment.]
