$\DeclareMathOperator{\Lim}{Lim}$Suppose the category $J$ is a coproduct $J = \amalg_k J_k$ with injections $I_j:J_j\to J$. For any functor $F:J\to \mathcal{C}$ this determines the component functors $F_k\equiv F\circ I_k$. Suppose furthermore that $\prod_k\Lim\ F_k$ exists in $\mathcal{C}$. The task is now to
Show that $\Lim F \cong \prod_k\Lim F_k$.
I am aware of this question, but both the question and the answer seem to assume that $\Lim F$ already exists, without proving it. My solution is the following:
- Note that we have morphisms $\prod_k\Lim F_k\xrightarrow{p_m}\Lim F_m \xrightarrow{\nu_m^i} F_mi= Fi$ s.t. $\nu^j_m = F_mu\circ \nu^i_m$ for all objects $i,j$ and morphisms $u:i\to j$ in $J$.
- Suppose that there exists an object $X$ and morphisms $x_i:X\to Fi$ s.t. $x_j = Fu\circ x_i = F_mu\circ x_i$ for all valid combinations of the symbols $m,i,j,u$. ($X$ is a cone over $F$)
- By the universal property of the limit $\exists!h_m:X\to\Lim F_m$ s.t. $x_i = \nu_m^i\circ h_m$ for all $i$.
- By the universal property of the product there then exists exactly one $h:X\rightarrow \prod_k\Lim F_k$ s.t. $h_m = p_m\circ h$ for all $m$
- Therefore, $\prod_k\Lim F_k$ is a limiting cone over $F$, i.e. a limit.
I am confused why we are asked to show the isomorphism $\Lim F\cong \prod_k \Lim F_k$. Did I do something wrong? Or is it simply the statement
If $\Lim F$ is already given, then we have just shown another way of obtaining it, since all limits are canonically isomorphic
?
Edit: This is question IV.2.7 (a) in Categories for the Working Mathematician (page 90 in the second edition)