TITLE: Limits interchange property in complete categories
QUESTION [5 upvotes]: Let $F:\mathscr{C}\times\mathscr{D}\longrightarrow\mathscr{A}$ be a functor, with $\mathscr{A}$ complete. I want to prove
$$\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))\cong\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D))$$
For every objet $C\in\mathscr{C}$, define
$$\lim_{D\in\mathscr{D}}F(C,D):=\lim F(C,-)$$
with projections $p^C_D:\lim_{D\in\mathscr{D}}F(C,D)\longrightarrow F(C,D)$
Fact: for every morphism $c:C\longrightarrow C'$ in $\mathscr{C}$, $(\displaystyle\lim_{D\in\mathscr{D}}F(C,D),(F(c,1_D)\circ p^C_D)_{D\in\mathscr{D}})$ is a cone over $F(C',-)$.
Thus there exists a unique factorization
$$\lim_{D\in\mathscr{D}}F(c,1_D):\lim_{D\in\mathscr{D}}F(C,D)\longrightarrow \lim_{D\in\mathscr{D}}F(C',D)$$
such that $p^{C'}_D\circ\displaystyle\lim_{D\in\mathscr{D}}F(c,1_D)=F(c,1_D)\circ p^C_D$. We may define a functor $L:\mathscr{C}\longrightarrow\mathscr{A}$, given by $LC=\displaystyle\lim_{D\in\mathscr{D}}F(C,D)$ and $Lc:=\displaystyle\lim_{D\in\mathscr{D}}F(c,1_D)$. We shall call
$$\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D)):=\lim L$$
the limit of this functor, with projections $p_C:\displaystyle\lim_{C\in\mathscr{C}}(\displaystyle\lim_{D\in\mathscr{D}}F(C,D))\longrightarrow\displaystyle\lim_{D\in\mathscr{D}}F(C,D)$.
Fact: $(\displaystyle\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D)),(p^C_D\circ p_C)_{C\in\mathscr{C}})$ is a cone over $F(-,D)$
Therefore there exists a unique factorization
$$\lambda_D:(\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))\longrightarrow\lim_{C\in\mathscr{C}}F(C,D)$$
such that $p^D_C\circ\lambda_D=p^C_D\circ p_C$.
Now, in a similar way, we may define for every object $D\in\mathscr{D}$
$$\lim_{C\in\mathscr{C}}F(C,D):=\lim F(-,D)$$
with projections $p^D_C:\displaystyle\lim_{C\in\mathscr{C}}F(C,D)\longrightarrow F(C,D)$.
Fact: for every morphism $d:D\longrightarrow D'$, $(\displaystyle\lim_{C\in\mathscr{C}}F(C,D), (F(1_C,d)\circ p^D_C)_{C\in\mathscr{C}})$ is a cone over $F(-,D')$.
So there exsts a unique factorization
$$\lim_{C\in\mathscr{C}}F(1_C,d):\lim_{C\in\mathscr{C}}F(C,D)\longrightarrow\lim_{C\in\mathscr{C}}F(C,D')$$
such that $p^{D'}_C\circ \displaystyle\lim_{C\in\mathscr{C}}F(1_C,d)=F(1_C,d)\circ p^D_C$. We may define a functor $M:\mathscr{D}\longrightarrow\mathscr{A}$ sending each object $D\in\mathscr{D}$ to $MD=\displaystyle\lim_{C\in\mathscr{C}}F(C,D)$ and each morphism $d:D\longrightarrow D'$ to $\displaystyle\lim_{C\in\mathscr{C}}F(1_C,d)$. We shall write
$$\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D)):=\lim M$$
for the limit of this functor, with projections
$$p_D:\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D))\longrightarrow\lim_{C\in\mathscr{C}}F(C,D)$$
Fact: $(\displaystyle\lim_{D\in\mathscr{D}}(\displaystyle\lim_{C\in\mathscr{C}}F(C,D)),(p^D_C\circ p_D)_{D\in\mathscr{D}})$ is a cone over $F(C;-)$
Thus we have a factorization
$$\mu_C:\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D))\longrightarrow\lim_{D\in\mathscr{D}}F(C,D)$$
such that $p^C_D\circ\mu_C=p^D_C\circ p_D$
Fact: $(\lambda_D)_{D\in\mathscr{D}}$ is a cone over $M$, and similarly $(\mu_C)_{C\in\mathscr{C}}$ is a cone over $L$.
Thus we have unique factorizations
$$\lambda:\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))\longrightarrow\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D))$$
and
$$\mu:\lim_{D\in\mathscr{D}}(\lim_{C\in\mathscr{C}}F(C,D))\longrightarrow\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))$$
such that $p_D\circ\lambda=\lambda_D$ and $p_C\circ\mu=\mu_C$
Proving the interchange property is equivalent to prove that $\lambda$ and $\mu$ are inverse isomorphisms. This is precisely what I can't prove. A proof of this fact can be found in "Handbook of categorical algebra, Vol.1", Borceux, at pag.74, but I can't still understand it.

REPLY [2 votes]: Note that a morphism with target $\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))$ is uniquely determined by the compositions with the projections $p_C : \lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D)) \to \lim_{D\in\mathscr{D}}F(C,D)$ for $C \in \mathscr{C}$. Again, for any $C \in \mathscr{C}$, a morphism to $\lim_{D\in\mathscr{D}}F(C,D)$ is uniquely determined by its compositions with the projections $p^C_D : \lim_{D\in\mathscr{D}}F(C,D) \to F(C,D)$. Hence a morphism to $\lim_{C\in\mathscr{C}}(\lim_{D\in\mathscr{D}}F(C,D))$ is uniquely determined by its compositions with $p^C_D \circ p_C$ for $C \in \mathscr{C}$ and $D \in \mathscr{D}$.
After having realized that, we compute $$p^C_D \circ p_C \circ (\mu \circ \lambda) = p^C_D \circ \mu_C \circ \lambda = p^D_C \circ p_D \circ \lambda = p^D_C \circ \lambda_D = p^C_D \circ p_C \circ \text{id}$$
and conclude that $\mu \circ \lambda = \text{id}$. Analogously you get that $\lambda \circ \mu = \text{id}$.