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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.

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Actually creating and reflecting limits are different things as shown in either this link or this other link. –  Giorgio Mossa Jan 6 at 19:43

2 Answers 2

up vote 0 down vote accepted

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.

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If $D$ is a diagram in $\cal D$, then $UD$ is a diagram in $\cal C$. By completeness, there is a limiting cone $(a,\ \alpha_k:a\to UDk)_{k\in J}$. Now $U$ creates limits, so there is a unique cone $(b,\ \beta_k:b\to Dk)_{k\in J}$, such that $Ub=a$ and $U\beta_k=\alpha_k$, and this cone is a limiting cone in $\cal D$. This shows that $\cal D$ is complete.
If $(b',\ \beta'_k:b'\to Dk)_{k\in J}$ is another limiting cone over $D$, then there is a isomorphism $f:b\to b'$ such that $β'_kf=β_k$. But then $Uf$ is an isomorphism $a\to Ub'$ such that $Uβ'_k\circ Uf=α_k$, which means that $(Ub', Uβ'_k:Ub'\to UDk)_{k\in J}$ is also a limiting cone over $UD$.

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