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Question: I want to define terminal objects from the categorical definition of limit, but I cannot. What is my mistake in the following argument?

Limit: Given a functor $F: \mathcal{D} \longrightarrow \mathcal{C}$, a limit of $F$ is a cone $L$ such that for every $M$, these following sets are naturally isomorphic $$\mathcal{C}(M, L) \cong \textbf{Nat}(\Delta_M, F)$$ ($\Delta_M$ is the constant functor and Nat is the set of natural transformation between these two functors $\Delta_M, F$)

Recovery of terminal object: Let the category $\mathcal{D}=\emptyset$ be the empty category, and $F=\Delta_{L}$. Then, for every $M$, $$\mathcal{C}(M, L) \cong \textbf{Nat}(\Delta_M, \Delta_{L})$$ $\Delta_L$ assignes $L$ to the empty category. So this isomorphy is trivial because both sets are the same. But it does not say $\mathcal{C}(M, L) $ has only one element, or $M \longrightarrow L$ is unique.

Where is my mistake?

Notice: The question here is regarding the categorical definition of limit and I am not asking for an explanation, I am looking for my mistake in my argument. (Response to a possible duplicate of other questions! )

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  • $\begingroup$ What's $\Delta_M$ in the context $D=\emptyset$ ? And thus, what is $Nat(\Delta_M,\Delta_\emptyset)$ ? $\endgroup$ – Maxime Ramzi Aug 21 '18 at 16:50
  • $\begingroup$ Edited and Added definitions. $\endgroup$ – Mobius Knot Aug 21 '18 at 17:23
  • $\begingroup$ No that was not the point of my question, my question was actually to hint you at a solution $\endgroup$ – Maxime Ramzi Aug 21 '18 at 17:27
  • $\begingroup$ This has been asked before, multiple times, including here. $\endgroup$ – Malice Vidrine Aug 21 '18 at 17:42
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    $\begingroup$ Possible duplicate of Explanation of terminal objects using limits $\endgroup$ – Malice Vidrine Aug 21 '18 at 17:43
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It's not clear what you mean by "$\Delta_L$ assigns $L$ to the empty category".

For any set $S$, there is a unique function $\varnothing \to S$. Using this function where appropriate, you can see:

  • There is a unique functor $E : \varnothing \to \mathcal{C}$
  • There is a unique natural transformation $\epsilon : E \to E$.

Consequently, $\textbf{Nat}(\Delta_M, \Delta_{L})$ is the one point set $\{ \epsilon \}$.

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  • $\begingroup$ That was my mistake. I did not notice that empty functor does not assign any objects or morphisms to the empty category. Its image is also empty. $\endgroup$ – Mobius Knot Aug 22 '18 at 8:51

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