Definition of category and natural transformations The book that I'am reading mentions the definition of a category as follows.
A category is a collection of abjects $\{X,Y, \ldots \}$ such that for two objects X, Y we have a set $\text{Mor}(X,Y)$ and for three objects $X, Y, Z$ a mapping (composition law),
$$ \text{Mor}(X,Y) \times \text{Mor}(Y,Z) \rightarrow \text{Mor}(X,Z)$$
satisfying the following axioms:


*

*Two sets $\text{Mor}(X,Y)$ and $\text{Mor}(X',Y')$ are disjoint unless $X=X'$ and $Y=Y'$, in which case they are equal.

*Each $\text{Mor}(X,X)$ has an element $id_X$.

*The composition law is associative.


Then a bit further in the text they define the natural transformation, which are the morphism of the category formed by the functors. If $\lambda, \mu$ are two functors from $\mathcal{U}$ to $\mathcal{U}'$ (say covariant), then a natural transformation $t:\lambda \rightarrow \mu$ consists of a collection of morphisms
$$ t_X : \lambda(X) \rightarrow \mu(X)$$
as $X$ ranges over $\mathcal{U}$, which makes the diagram commutative for every $f: X\rightarrow Y$, it is,
$$ \mu(f) \circ t_X = t_Y \circ \lambda(f).$$
My question is how one can show that this category indeed does satisfy the first point in the definition? (By this category I mean the category formed by the functors and where the morphisms are the natural transformations)
My attempt:
Consider that the two sets $\text{Mor}(\lambda,\mu)$ and $\text{Mor}(\lambda',\mu')$ are not disjoint. Then I would like to show that $\lambda=\lambda'$ and $\mu=\mu'$. Due to the consideration is it possible to take $t$ in the section of $\text{Mor}(\lambda,\mu)$ and $\text{Mor}(\lambda',\mu')$. Take an arbitrary set $X\in\mathcal{U}$ then,
$$ t_X \in \text{Mor}(\lambda(X),\mu(X)) \\ t_X \in \text{Mor}(\lambda'(X),\mu'(X)) $$
Due to the first point there follows that $\lambda(X) = \lambda'(X)$ and $\mu(X) = \mu'(X)$. And since $X$ was arbitrary this will be the case for every $X\in\mathcal{U}$.
It is at this point that I got stuck, could someone please help me further? Thanks in advance.
 A: You can't really prove that the actual collections of morphisms that constitute a natural transformation between two functors are distinct, because it's not always the case.
For example, if $\mathcal{U'}$ is the category $\mathbf{Vect}_{\Bbb R}$ of real vector spaces, then for any functor $\mu:\mathcal{U}\to \mathbf{Vect}_{\Bbb R}$ and any $a\in \mathbb{R}$, you have a natural transformation $\mu\Rightarrow \mu$ defined by $t_X:\mu(X)\to \mu(X):v\mapsto av$. This is a natural transformation regardless of how $\mu$ is defined on arrows, so if $\mu'$ is a functor $\mathbf{U}\to \mathbf{Vect}_{\Bbb R}$ that agrees with $\mu$ on objects (i.e. such that $\mu(X)=\mu'(X)$ for all $X$ of $\mathcal{U}$) but not on morphisms, then the same collection $(t_X:\mu'(X)\to \mu'(X))$ would also constitute a natural transformation $\mu'\Rightarrow \mu'$.
In fact, a part of this example works for any category $\mathcal{U'}$ : the identity natural transformation $\mu\Rightarrow \mu$ is always given by the collection of identities
$$(id_{\mu_X}:\mu(X)\to \mu(X))_{X\in Ob (\mathcal{U})},$$
so it does not depend on how $\mu$ is defined for morphisms. Thus if $\mu'$ is as above, we have again that the identity $id_\mu$ is technically equal to the identity $id_{\mu'}$. The example above is basically the same case, except that since the category $\mathbf{Vect_\Bbb{R}}$ is enriched over itself, all multiples of the counterexample are also counterexamples.

The point is, as Alex Provost said in his answer, that we consider that every natural transformation $t:\lambda\Rightarrow \mu$ has an assigned domain $\lambda$ and codomain $\mu$, and that these are part of the definition of $t$. This is similar to the case of functions between sets : if you define a function $f:A\to B$ as a set $\Gamma \subset A\times B$ such that for every $a\in A$ there is a unique pair $(a,b)\in \Gamma$, and then two functions with different codomain can technically be seen as the same set $\Gamma$. But if you consider that $A$ and $B$ are part of the definition of $f$, this is not possible.
A: The purpose of this post is to point out what I believe is a mistake in the question and in the accepted answer. 
The OP defines a category as 

a collection of objects $\{X,Y, \ldots \}$ such that for two objects X, Y we have a set [emphasis added] $\text{Mor}(X,Y)$ ... 

Later the OP writes

... the natural transformations, which are the morphism of the category [emphasis added] formed by the functors ...

The statement that the collection of morphisms between two functors is a set is made repeatedly throughout the thread.
Let's show that it is not so:
Let $\mathcal C$ be the unique category whose objects are the sets and whose morphisms are defined as follows: $\hom_{\mathcal C}(X,Y)=\varnothing$ if $X\neq Y$ and $\hom_{\mathcal C}(X,X)=\{\text{id}_X\}$.
Let $\mathcal C'$ be the usual category of sets.
Let $F$ be the unique functor from $\mathcal C$ to $\mathcal C'$ such that $F(X)=X$ for all $X$.
If there was a set $S$ whose elements are the endomorphisms of $F$, then for each set $X$ we would have a surjection from $S$ onto the set of endomaps of $X$. This is clearly impossible.
Edit. Let us define a category as given by a collection of objects, and, for each pair of objects, a collection of morphisms satisfying the usual axioms, without the disjointness condition. Then it is easy to see that each category is canonically isomorphic to a category with disjoint Hom-collections.
A: There is not much to prove here. By definition, a natural transformation $t:\lambda \to \mu$ goes from a specified functor $\lambda$ to another specified functor $\mu$. That is, the "source-target" data $(\lambda,\mu)$ is uniquely determined by $t$. A fortiori, a single natural transformation cannot have two distinct sources or targets, which implies disjointness of the morphism sets.
