From pg. 56 of Categories for the Working Mathematician:

Definition. If $S:D\rightarrow C$ is a functor and $c$ and object of $C$, a universal arrow ... $u: c \rightarrow Sr$ is universal from $c$ to $S$ when the pair $\langle r, u \rangle$ is an initial object in the comma category $(c \downarrow S)$, whose objects are the arrows $c \rightarrow Sd$. As with any initial object, it follows that $\langle r, u \rangle$ is unique up to isomorphism in $(c \downarrow S)$, in particular, the object $r$ of $D$ is unique up to isomorphism in $D$.

Question: In the textbook, this bolded part is asserted without argument. Why is it true?

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    $\begingroup$ Are you happy with the fact that $\langle r,u \rangle$ is unique up to isomorphism in the comma category? $\endgroup$ – Patrick Stevens Mar 8 '17 at 10:05
  • $\begingroup$ Yes, I understand that. $\endgroup$ – user1770201 Mar 8 '17 at 10:05
  • $\begingroup$ (I'll leave it to someone else to answer, I'm afraid, because one of my hands is unusable. Makes typing difficult :P ) $\endgroup$ – Patrick Stevens Mar 8 '17 at 10:10

Assume $r'$ is another 'such' object in $D$.

What could it mean in this context? That there is an arrow $u':c\to Sr'$ such that $\langle r',u'\rangle$ is an initial object of the comma category.

Since $\langle r,u\rangle$ is also initial, they are isomorphic in the comma category, meaning that there is an isomorphism $d\in D$ making the commutative triangle $Sd\circ u = u'$.

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    $\begingroup$ I think it clarifies this answer to mention that the morphisms $d:r\to r'$ and $d':r'\to r$ are inverses because initiality here means there is exactly one morphism $f:r\to r$ with $Sf\circ u=u$, and this is obviously true of $id_r$; hence if it's also true of $d'\circ d$, then this must equal $id_r$. (And likewise for $r'$.) I mention this because this was the detail that was opaque to me the first time I encountered this theorem, so maybe it's helpful here. $\endgroup$ – Malice Vidrine Mar 9 '17 at 3:25

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