In the remark author says:

A. there is a function U:Ob(Q)→Ob(C),

B. for all a,b∈Ob(Q), we have an injection U:HomQ(a,b)↪HomC(U(a),U(b)),

Why is the function and injection both called U ? Shouldn’t the injection be called something other than U ? This is more a question of naming but I just wanted to confirm.

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Good point!

What $A$ shows is that $U$ is a function.

What $B$ shows is that $U$ preserves morphisms and that too injectively which becomes quite handy when we want to prove (1,2).

As an example, take Groups. Group elements can be represented as invertible matrices which form a subgroup of General Linear Group -. $GL(V)$. Thus, we have group morphism or homomorphism and a map.

Then a function similar to $U$ called $\rho$ maps each group element $g$ to $\rho(g)$. And for $g_1,g_2 \in G$, we have matrices $\rho(g_1),\rho(g2)$. Since $g_3=g_1*g_2$, to preserve group structure, we need $\rho(g_1*g_2) = \rho(g_1)*\rho(g_2)=\rho(g_3)$. This is same as $B$ but also includes additional requirement – injection.

How does the injection help?

Take questionable Category $Q$. Then attempt to prove $1$ which is identity morphism in the domain and co-domain.We need to prove $f \circ id_x = f$ and $id_y \circ f = f$. To prove $f \circ id_x = f$ in $Q$, we need to move this equation to known category $C$ using $U$.

Thus,

$U(f \circ id_x) = U(f) \circ U((id_x)

= U(f) \circ id_{U(x)}

= U(f)$

Because of injectivity, we can assert $f \circ id_x = f$.

I understood the idea there. We need objects a, b and functions a -> b both to go from Q to C. I was talking more about the notation. Why are both the function and injection named U ? Shouldn’t they be named differently ? Does U encompass both the morphism of objects and functions ? That makes U a functor(which has not been introduced at this point). I think they should be named differently.

I believe you are right on this. The naming convention is introduced later in Definition 5.1.2.1 when talking about functors. A functor F between two categories includes a mapping between objects and a mapping between morphisms, and they are both denoted F when this is unambiguous.

In fact you are correct, U is exactly a functor if Q is an actual category. Of course but the remark states that the existence of U guarantees that Q is a category.