Proving that $p_I\circ h=p_I\circ h'$ implies $h=h'$

Question

An exercise from Leinster:

I was wondering if my solution is correct?

(a) Assume $h,h':A\to L$ are arrows such that $p_I\circ h=p_I\circ h'$ for all $I$. Define $f_I=p_I\circ h$, as shown on the picture. Then $(f_I:A\to D(I))_{I\in\mathbf I}$ is a cone on $D$. Since there is only one map that makes the orange triangle commute (for all $I$), and since both $h$ and $h'$ make the diagram commute (since $p_I\circ h=p_I\circ h')$, it follows that $h=h'$.

(b) Let $1=\{\star\}$, let the objects of the discrete category be $S,T$. We have the diagram

The given condition says that the equations $$p_1(h(\star))=p_1(h'(\star))\\ p_2(h(\star))=p_2(h'(\star))$$

imply $h=h'$.

Is that was is asked? Or do I need to say more? If so, what exactly?

Answer 1

a) is correct

b) This limit should be familiar as the product, so $L = S \times T$, morphisms from the initial object $1 \to X$ can be thought of as elements of $X$, and then the condition in a) reads: Let $h = (s, t)$ and $h' = (s', t')$. If $s = p_1 \circ h = p_1 \circ h' = s'$ and $t = p_2 \circ h = p_2 \circ h'= t'$, then $h = h'$.