equality of two natural transformations and two morphisms When are two natural transformations between two functors considered the same?  When are two morphisms in a category considered the same? Do we have such a notion like we have in Set theory? 
Thanks
 A: As @Zhen Lin and others noted, the notion of equality is sometimes debated.
I personally like the following approach since it is quite general.
In the constructive school of thought, a category is a type of objects, arrows, composition,
and identities satisfying certain laws, but moreover it is also equipped with an equivalence
relation on arrows of the same type: given $f, g : A ⟶ B$ morphisms in the category, we can
form the expression $f ≈ g$. Notice that it need not be strict equality but any equivalence
relation, and it is on the morphisms not the objects.
Then the problem of morphism equality is solved by the implementer, the one who defined the
equality, and so the user (of the category) need not worry about it (or about having to make a
choice of when two things are 'equal'). For example, the morphism-equivalence in one category
might be an isomorphism in simpler category!
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The reason that there is no corresponding notion of equality of objects, in the constructivist approach, is that categories are all about morphisms not objects and usually one uses object-isomorphism anyways.
⟧
Time for an example. Suppose we have two categories  and , with morphism-equalities ≈₀ and ≈₁, then
the functor category $Func \;  \; = [, ] =  ^ $ may be defined to have morphism (ie natural
transformation) equality to be pointwise:
$γ ≈ η \;⇔\; (∀ x • γ_x \;≈_1\; η_x)$ ---notice that we're using the equality of -morphisms.
Another example: The category of categories  has equality of functors ≈ being
pointwise equalities $$F ≈ G :  ⟶  \;⇔\; (∀ x : Obj \, • F x = G x)$$ $$∀ f, g : Arrow •\; f \; ≈_0 \; g \;\;⇒\;\; F f \;≈_1\; F g$$
Where ,  have morphism equalities ≈₀ and ≈₁, respectively.
Alternatively, one may define $F ≈ G ⇔ (∀ x • F x ≅ G x \text{ in  and naturally so in $x$})$, i.e.,
functor-isomorphism and then the functor category, is by this particular construction, skeletal(!): all isomorphisms
are equality. Why you want this is another matter.
Finally note that a common construction is to define an equivalence relation on morphisms
then quotient by it to obtain equivalence classes and then consider this category, each time
having to prove well-definedness due to the usage of representatives. This now becomes
by proving operations are congruences.
Hope this helps!
A: Actually this question is equivalent to the question "When are two elements of a set considered same?" The answer is "When they are equal".
Indeed, let $C$ be a category, $f,g\in Arr(C)$ be its morphisms. Then $f=g$ as morphisms iff $f=g$ as elements of the set $Arr(C)$.
Let $C$ and $D$ be categories, $T,S\colon C\to D$ be functors, $\alpha,\beta\colon T\to S$ be natural transformations. Then $\alpha=\beta$ as natural transformations iff $\alpha(c)=\beta(c)$ for every $c\in Ob(C)$. Equivalently, $\alpha=\beta$ as natural transformations iff $\alpha=\beta$ as elements of $Nat(T,S)$ - the set of natural transformations between $T$ and $S$. This set consists of mappings $Ob(C)\to Arr(D)$ with additional properties (subsets of $Ob(C)\times Arr(D)$ with additional properties), so its elements are equal when they are equal as mappings (as subsets of $Ob(C)\times Arr(D)$).
