An existence of a right Kan extension Let $M$, $C$ and $A$ be categories. The following statement is well known.

If $M$ is small and $A$ is complete, then any functor $T:M \to A$ has
  a right Kan extension along any $K:M \to C$.

This statement can be found in page 239 of "Categories for Working Mathematician" (by Mac Lane). In this book, the statement is a corollary of a theorem saying that a right Kan extension can be obtained as a limit for a certain functor $(c \downarrow K) \to A$ whenever the limits exist for all object $c$ in $C$.
My question is: why is there no need to assume that $C$ is locally small?
If $A$ is complete, then any small diagrams in $A$ have limits in $A$ by definition. Thus we want $(c \downarrow K)$ to be small. But, for an object $m$ of $M$, the collection of arrows $\{\alpha:c\to K(m)\}$ might not be a set, unless $C$ is locally small. How can we avoid this problem?
 A: Let $\mathcal{M} = \mathbb{1}$ be the terminal category, let $\mathcal{A}$ be a complete locally small category, and let $\mathcal{C}$ be a category with two objects $c$, $d$, such that the only non-trivial morphisms in $\mathcal{C}$ are those $c \to d$. A functor out of $\mathcal{M}$ is the same thing as choosing an object, so there is a functor $K : \mathcal{M} \to \mathcal{C}$ with image $d$. Suppose $\mathcal{A}$ has an object $a$ such that, for some $b$ in $\mathcal{A}$, there exist at least two distinct morphisms $b \to a$. Let $T : \mathcal{M} \to \mathcal{A}$ be the functor picking out $a$. 
Suppose $\mathcal{C}(c, d)$ is a proper class. I claim the right Kan extension of $T$ along $K$, if it exists, is not computed by the quoted limit. Indeed, note that the comma category $(c \downarrow K)$ is a large discrete category (i.e. there is a proper class of objects but all morphisms are trivial), so the limit of 
$$(c \downarrow K) \to \mathcal{M} \to \mathcal{A}$$
must therefore be the product of $\mathcal{C}(c, d)$-many copies of $a$. However, no such product can exist: if $e$ were such a product, then 
$$\mathcal{A}(b, e) \cong \mathcal{A}(b, a)^{\mathcal{C}(c, d)}$$
but the RHS is a proper class, since $\mathcal{A}(b, a)$ has at least two elements and $\mathcal{A}$ is locally small.
This does not preclude the possibility that the right Kan extension exists for other reasons – but if it does, then we know very little about it because it is not computed by the usual means. Thus, we must conclude that the quoted theorem requires $\mathcal{C}$ to be locally small, or at the very least that $\mathcal{A}$ has limits for some large diagrams.
