First, I will fix a relevant definition. Given a small category $\mathsf{I}$, a diagram $\mathcal{D}\colon\mathsf{I}\to\mathsf{C}$, a functor $F\colon\mathsf{C}\to\mathsf{D}$ and a cone $\lambda\colon X\Rightarrow \mathcal{D}$, there is a cone $F(\lambda)\colon F(X) \Rightarrow F\mathcal{D}$ defines as follows: for any $i \in \mathsf{I}$, $F(\lambda)_i = F(\lambda_i)$. It is straightforward to make sure that this data indeed defines a cone. Such constructions are important in theory of functors which preserve, reflect or create limits or colimits.
Let $\mathsf{I}$ be a small category, let $A$ be a set and let $\mathsf{C}$ be a category. Let $\prod_{A}\mathsf{C}$ be a product category of the family $(\mathsf{C})_A$ (the family $(\mathsf{C}_a)_{a \in A}$ so that for any $a \in A$ we have $\mathsf{C}_a = \mathsf{C}$). Let $(\Pi_a\colon\prod_{A}\mathsf{C}\to\mathsf{C})_{a \in A}$ be a family of projection functors:
for any $(X_a)_{a \in A}$, $\Pi_a((X_a)_{a \in A}) = X_a$,
for any $(f_a\colon X_a \to Y_a)_{a \in A}$, $\Pi_a((f_a)_{a \in A}) = f_a$.
Let $\mathcal{D}\colon\mathsf{I}\to\prod_{A}\mathsf{C}$ be a diagram and let $\lambda\colon (l_a)_{a \in A} \Rightarrow \mathcal{D}$ be a limit cone. Then, for any $a \in A$, $\Pi_a(\lambda)\colon l_a \Rightarrow \Pi_a\mathcal{D}$ is also a limit cone.
I want to make sure that this assertion and the following proof of it are right.
Let $a' \in A$. Let $\mu\colon Y \Rightarrow \Pi_{a'}\mathcal{D}$ be a cone. Consider a family $(Y_a)_{a \in A}$ so that, for any $a \in A$, $Y_a$ is equal to $Y$ if $a = a'$, and is equal to $l_a$ otherwise. Consider a hypothetical cone $\nu\colon (Y_a)_{a \in A} \Rightarrow \mathcal{D}$ so that, for any $a \in A$, $\nu_a$ is equal to $\mu$ if $a = a'$, and is equal to $\Pi_a\lambda$ otherwise. We claim that $\nu$ is indeed a cone. Let $f\colon i\to j$ be a morphism of $\mathsf{I}$. As $\mu$ is a cone, the we have $\Pi_{a'}(\mathcal{D}(f)) \circ \mu_i = \mu_j$. As each $\Pi_a(\lambda)$ is a cone, then, for any $a \in A\setminus\{a'\}$, we have $\Pi_a(\mathcal{D}(f))\circ\Pi_a(\lambda_i) = \Pi_a(\lambda_j)$. As $\mathcal{D}(f) = (\Pi_a(\mathcal{D}(f)))_{a \in A}$ and considering the definition of $\nu$, we thus have $\mathcal{D}(f)\circ\nu_i = \nu_j$. Hence, the data defining $\nu$ in fact defines a cone.
Since $\lambda$ is a limit cone, there is a unique morphism $(f_a)_{a \in A}\colon (Y_a)_{a \in A} \to (l_a)_{a \in A}$ so that for any $i \in \mathsf{I}$ we have $\lambda_i\circ (f_a)_{a \in A} = \nu_i$. In particular, it means that $\Pi_a(\lambda_i)\circ f_{a'} = \mu_i$ for any $i \in \mathsf{I}$. Let $g\colon Y\to l_{a'}$ be another morphism for which we have $\Pi_a(\lambda_i)\circ g = \mu_i$ for any $i \in \mathsf{I}$. Consider a family $(g_a\colon Y_a\to l_a)$ so that $g_a = g$ is $a = a'$, and $g_a = f_a$ otherwise. It is clear that we then have $\lambda_i\circ (g_a)_{a \in A} = \nu_i$ for any $i \in \mathsf{I}$, which, in turn, implies that $(f_a)_{a \in A} = (g_a)_{a \in A}$, hence $g = g_{a'} = f_{a'}$.