In Saunders Mac Lane's Categories for the working mathematician one can read, when talking about fully faithful functors:

[...], but this need not mean that the functor itself is an isomorphism of categories, for there may be objects of B not in the image of T

Given a fully faithful functor $T: C \to B$ and an object together with its identity arrow $c \in C, 1_c: c \to c$, and given $T 1_c = 1_{Tc}: T c \to T c$ in $B$, how can there be a fully faithful functor between categories with unequal amounts of elements?

Because faithful functors follow the rule $Tf_1 = Tf_2 \Rightarrow f_1 = f_2$ and with categories with unequal (or rather less) elements there must be some $1_{Tc}$ that is equal to some $1_{Tc'}$

For me it seems to be the same reason why there is no injective function $\mathbb{Z} \to \mathbb{N}$

  • 1
    $\begingroup$ For any functor $T1_c=1_{Tc}$. $\endgroup$ Oct 1 '17 at 18:06
  • 1
    $\begingroup$ Functors map identity morphisms to identity morphisms by definition. $\endgroup$
    – Arthur
    Oct 1 '17 at 18:08
  • 1
    $\begingroup$ Further to the above comments, here's a silly class of examples of full and faithful functors between categories with unequal numbers of objects. Let $\mathcal{C}$ be any indiscrete category, i.e. a set of objects such that between any two objects there is exactly one morphism, and let $\mathbf{1}$ be the terminal category, i.e. the category with one object and one (identity) morphism. The unique functor $\mathcal{C} \to \mathbf{1}$ is full and faithful. $\endgroup$ Oct 1 '17 at 18:12
  • $\begingroup$ @CliveNewstead Really? A faithful functor implies $Tf_1 = Tf_2 \Rightarrow f_1 = f_2$, but if $\forall f \in \mathcal{C}. T f = id_1$, (where $id_1$ is the single arrow in the terminal category) then this doesn't hold. $\endgroup$
    – hgiesel
    Oct 1 '17 at 20:24
  • 1
    $\begingroup$ Because $Tf_1= Tf_2 \implies f_1=f_2$ need not hold, even if $T$ is faithful. For it to hold, you have to assume first that $f_1, f_2$ have the same domain and codomain: a faithful functor is injective when restricted to $Hom(A,B)$, not when restricted to $Mor(C)$ $\endgroup$ Oct 1 '17 at 20:55

Let $\mathcal C$ be any full subcategory of a category $\mathcal D$. Then the inclusion $\mathcal C \subseteq \mathcal D$ gives a functor $F\colon\mathcal C \to \mathcal D$. This functor is always fully faithfull, but it is not always true that a subcategory is isomorphic to the larger category.

For a really down to earth example take the category whose only object is the zero set with its identity map. This includes into the category of all sets and these categories are not isomorphic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.