[proofplan] We prove preservation by comparing an arbitrary limiting cone over $D$ with the limiting cone obtained by lifting a limiting cone over $F\circ D$ through the creation property. Since limits are unique up to unique isomorphism, the image of the original limiting cone is isomorphic as a cone to a limiting cone in $\mathcal D$, hence is itself limiting. Reflection is more direct: if a cone over $D$ maps to a limiting cone over $F\circ D$, then the creation property says that the unique lift of that image cone is limiting; the given cone is already such a lift, so it must be that limiting cone. [/proofplan] [step:Fix the notation for cones and the creation hypothesis] For an object $A\in\mathcal C$, write $\Delta A:J\to\mathcal C$ for the constant diagram at $A$. A cone over $D$ with vertex $A$ is a natural transformation \begin{align*} \alpha:\Delta A\Rightarrow D. \end{align*} Its image under $F$ is the cone \begin{align*} F\alpha:\Delta F(A)\Rightarrow F\circ D \end{align*} whose component at $j\in J$ is $F(\alpha_j):F(A)\to F(D(j))$. We use the strict lifting convention for creation: the lifted vertex and cone have literal images, not merely specified isomorphic images. Thus the hypothesis that $F$ creates the limit of $D$ means the following. If \begin{align*} \rho:\Delta Y\Rightarrow F\circ D \end{align*} is a limiting cone in $\mathcal D$, then there exist an object $X\in\mathcal C$ and a cone \begin{align*} \widetilde\rho:\Delta X\Rightarrow D \end{align*} such that $F(X)=Y$ and $F\widetilde\rho=\rho$; this lifted cone is limiting in $\mathcal C$, and it is unique among cones over $D$ whose image is $\rho$. [/step] [step:Show that $F$ preserves the limiting cone] Let \begin{align*} \lambda:\Delta L\Rightarrow D \end{align*} be a limiting cone in $\mathcal C$. By the added hypothesis that $F\circ D$ has a limiting cone, choose a limiting cone \begin{align*} \rho:\Delta Y\Rightarrow F\circ D \end{align*} in $\mathcal D$. Since $F$ creates the limit of $D$, $\rho$ has a unique lifted cone \begin{align*} \widetilde\rho:\Delta X\Rightarrow D \end{align*} with $F(X)=Y$, $F\widetilde\rho=\rho$, and $\widetilde\rho$ limiting in $\mathcal C$. Now $\lambda$ and $\widetilde\rho$ are both limiting cones over $D$. Therefore there is a unique isomorphism \begin{align*} u:L\to X \end{align*} in $\mathcal C$ such that, for every object $j\in J$, \begin{align*} \widetilde\rho_j\circ u=\lambda_j. \end{align*} Applying $F$ gives, for every $j\in J$, \begin{align*} \rho_j\circ F(u)=F(\lambda_j). \end{align*} Thus $F(u):F(L)\to Y$ is an isomorphism of cones from $F\lambda$ to $\rho$. Since $\rho$ is a limiting cone in $\mathcal D$ and $F\lambda$ is isomorphic to $\rho$ as a cone over $F\circ D$, the cone $F\lambda$ is also limiting. Hence $F$ preserves the limit of $D$. [guided] Let \begin{align*} \lambda:\Delta L\Rightarrow D \end{align*} be a limiting cone in $\mathcal C$. To prove preservation, we must show that the image cone \begin{align*} F\lambda:\Delta F(L)\Rightarrow F\circ D \end{align*} is limiting in $\mathcal D$. The additional hypothesis that $F\circ D$ has a limit gives us a reference limiting cone in $\mathcal D$. Namely, choose a limiting cone \begin{align*} \rho:\Delta Y\Rightarrow F\circ D. \end{align*} Because $F$ creates the limit of $D$, this cone has a unique lift \begin{align*} \widetilde\rho:\Delta X\Rightarrow D \end{align*} with $F(X)=Y$ and $F\widetilde\rho=\rho$, and the lifted cone $\widetilde\rho$ is limiting in $\mathcal C$. Now we compare $\lambda$ with $\widetilde\rho$ inside $\mathcal C$. Both are limiting cones over the same diagram $D:J\to\mathcal C$. By the universal property of the limiting cone $\widetilde\rho$, there is a unique morphism \begin{align*} u:L\to X \end{align*} such that \begin{align*} \widetilde\rho_j\circ u=\lambda_j \end{align*} for every object $j\in J$. Since $\lambda$ is also limiting, the same universal property in the opposite direction supplies an inverse morphism, so $u$ is an isomorphism of cone vertices. Applying the functor $F$ to the cone identities gives \begin{align*} F(\widetilde\rho_j)\circ F(u)=F(\lambda_j). \end{align*} Since $F\widetilde\rho=\rho$, this becomes \begin{align*} \rho_j\circ F(u)=F(\lambda_j) \end{align*} for every $j\in J$. Thus $F(u):F(L)\to Y$ is an isomorphism of cones from $F\lambda$ to $\rho$. A cone is limiting exactly when it satisfies the universal property, and that property is invariant under isomorphism of cones. Since $\rho$ is limiting and $F\lambda$ is isomorphic to $\rho$ as a cone over $F\circ D$, the cone $F\lambda$ is limiting. This proves that $F$ preserves the limit of $D$. [/guided] [/step] [step:Show that $F$ reflects limiting cones] Let \begin{align*} \mu:\Delta X\Rightarrow D \end{align*} be a cone in $\mathcal C$ such that \begin{align*} F\mu:\Delta F(X)\Rightarrow F\circ D \end{align*} is limiting in $\mathcal D$. Since $F$ creates the limit of $D$, the limiting cone $F\mu$ has a unique lift to a cone over $D$, and that lift is limiting in $\mathcal C$. But $\mu$ is itself a lift of $F\mu$, because its image under $F$ is exactly $F\mu$. By uniqueness of the lift in the creation property, $\mu$ is the lifted limiting cone. Therefore $\mu$ is limiting in $\mathcal C$. This proves that $F$ reflects the limit of $D$. [/step] [step:Conclude preservation and reflection] The first argument shows that every limiting cone over $D$ is sent by $F$ to a limiting cone over $F\circ D$, so $F$ preserves the limit of $D$. The second argument shows that every cone over $D$ whose image under $F$ is limiting must itself be limiting, so $F$ reflects the limit of $D$. Hence created limits are both preserved and reflected. [/step]