I'm having trouble understanding the proof of the theorem on bifunctors in MacLane's CWM on page 37.
Let $\mathcal{C}$, $\mathcal{D}$ and $\mathcal{E}$ be categories and let $ L_y : \mathcal{C} \rightarrow \mathcal{E} $ and $ M_x : \mathcal{D} \rightarrow \mathcal{E} $ be parametrized functors where $ x \in \mathcal{C} $ and $ y \in \mathcal{D} $ such that $ L_y(x) = M_x(y) $ for all $x$ and $y$.
Then there exists a bifunctor $ F : \mathcal{C} \times \mathcal{D} \rightarrow \mathcal{E} $ such that $ F(-,y) = L_y $ and $ F(x,-) = M_x $ for all $x$ and $y$, iff for all pairs of morphisms $ f : c \rightarrow c' $ and $ g : d \rightarrow d' $ of resp. $ \mathcal{C} $ and $ \mathcal{D} $ we have $$ L_{d'}(f) \circ M_c(g) = M_{c'}(g) \circ L_d(f) $$
The proof of the converse is easy, but I feel like I'm missing something on the direct one :
Since we have $ \langle f, 1_{d'} \rangle \circ \langle 1_c, g \rangle = \langle 1_{c'}, g \rangle \circ \langle f, 1_d \rangle $, we get the following commutative diagram :
$$ \require{AMScd}\begin{CD} F(c,d) @>F(1_c,g)>> F(c,d')\\ @VF(f,1_d)VV @VVF(f,1_{d'})V\\ F(c',d) @>>F(1_{c'},g)> F(c',d')\end{CD} $$
But how can we say that $ F(1_c,-) $ is the arrow function of the $ M_c $ functor? (likewise for $ F(-, 1_d) $ and $ L_d $)
I believe Vladimir Sotirov's comment is correct: $F(c,–)=M_c$ is meant to denote that both $F(c,d)=M_cd$ for all objects $d$ of $\mathcal D$ and $F(1_c,f)=M_cf$ for all arrows $f$ of $\mathcal D$.
This is a flaw in the flow of the proof. The convention “we write b and c for the corresponding identity arrows” is only introduced in the proof, but it’s actually already used in the statement of the proposition, as $F(-,y)=L_y$ and $F(x,–)= M_x$ wouldn't make sense if $x$ and $y$ were really just the objects they’re introduced as: The right-hand sides are functors, so the left-hand sides must be able to map arrows, not just objects.