Exercise 5.3.10 from Leinster:
Prove that if a functor creates limits, then it also reflects them.
Here's what I have so far (the question is at the end):
Let $F:\mathscr A\to \mathscr B$ be a functor which creates limits and let $D:I\to \mathscr A$ be a diagram (=functor). Consider a cone $(A, f_i:A\to D(i))_{i\in I}$. Assume that its image under $F$, namely $(F(A), F(f_i):A\to FD(i))_{i\in I}$, is a limit cone on $F\circ D$. Need to prove that $(A, f_i:A\to D(i))_{i\in I}$ is a limit cone on $D$.
Since $F$ creates limits, there is a cone $(L, p_i:L\to D(i))_{i\in I}$ on $D$ such that $F(p_i)=F(f_i), F(L)=F(A)$, and this cone is a limit cone on $D$. It follows that there is a unique arrow $\alpha: A\to L$ such that $p_i\circ \alpha= f_i$ for all $i\in I$. So for any $i, j$ for which there is an arrow $i\to j$ in $I$, we have the following picture (on the left):
After taking $F$, we get the picture on the right. That picture implies that $F(\alpha)$ is the identity arrow. How to proceed? If we knew that $A=L$ then we could conclude that $\alpha=id$ (since $id$ must go to $id$, and $id$ is the only arrow from $A$ to $L$; but we wouldn't even need that since $A$ would be the limit automatically).
Best Answer
There is no need to draw any diagrams for this question. By definition of creating limits in Leinster (see Remark 5.3.7), any cone which maps to the limit cone must itself be a limit cone. Since $(A, f_i)_{i \in I}$ is certainly a cone which does so, it must be a limit cone and $F$ reflects limits.
If you want to use the strict definition (Def 5.3.5), it follows from uniqueness of the cone that $(L, p_i)_{i \in I}$ is the same cone as $(A, f_i)_{i \in I}$, and the result follows.