TITLE: Creates limits and preserve limits
QUESTION [5 upvotes]: If we have given that U is a functor  $ D \to C $ creates limits and C is Complete. and we are to show that D is complete and U preserves limits. 
I follow its in this way
As we know that U is a functor  $ D \to C $ creates limits if and only if U reflects limits that means all the diagrams D $ I \to D $ and all cones ($ q_X $ : $ A \to DX  )$  where $X \in I$ on D. if  $ Uq_X  :  UA \to UDX $   is a limit cone on D. This implies that UD has a limit in C, hence C has all limits. so it is complete.
However, I don't understand how we can prove D is complete by using this definition, if it satisfies above.

REPLY [2 votes]: The statement $U \colon \mathbf D \to \mathbf C$ create limits means that for every functor/diagram $F \colon \mathbf J \to \mathbf D$ for every limit cone $\langle \tau_j \colon d \to U \circ F(j)\rangle_{j \in \mathbf J}$ there's a unique limit cone $\langle \sigma_j \colon c \to F(j)\rangle_{j \in \mathbf J}$ such that $U(\sigma_j)=\tau_j$ for every $j \in \mathbf J$.
If $\mathbf C$ was complete (I suppose you mean small complete) then for every category (small) $\mathbf J$ and functor $F' \colon \mathbf J \to \mathbf C$ we have a limit cone $\langle \tau_j \colon d \to F'(j)\rangle_{j \in \mathbf J}$.
This in particular implies that for every functor $F \colon \mathbf J \to \mathbf D$ the functor $U \circ F \colon \mathbf J \to \mathbf C$ has a limit cone $\langle \tau_j \colon d \to U\circ F(j)\rangle_{j \in \mathbf J}$. By the property of $U$ of creating limit there must exists also a limit cone $\langle \sigma_j \colon c \to F(j)\rangle_{j \in \mathbf J}$: hence $F$ has a limit in $\mathbf D$. 
To prove that $U$ preserve limits let suppose that $\langle \sigma_j \colon c \to F(j)\rangle_{j \in \mathbf J}$ is a limit cone in $\mathbf C$ for $F$.
If $\langle \tau_j \colon d \to U\circ F(j)\rangle_{j \in \mathbf J}$ was limit cone for $U \circ F$ then by what we have seen above there's $\langle \sigma'_j \colon c' \to F(j)\rangle_{j \in \mathbf J}$ which is a limit for $F$.
But since also $\sigma$ was a limit cone for $F$ there must be an isomorphism $i \colon c \to c'$ in $\mathbf C$ that is a isomorphism of cones $f \colon \sigma \to \sigma'$ (i.e. for every $j \in \mathbf J$ we have that $\sigma'_j \circ i = \sigma_j$). $U(i)$ must is still an isomorphism between the cones $U(\sigma_j)$ and $U(\sigma')=\tau$, and since $\tau$ was a limit cone then also $U(\sigma)$ is a limit cone for $U \circ F$. So $U$ preserve limits.
Hope this helps.