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I am trying learn category theory from a programmer's perspective. Thus the following may be plain wrong. I want to know if my proof below is OK:

Show that the terminal object is unique upto a unique isomorphism

"Unique up to a unique isomorphism" implies the following: Unique up to unique isomorphism means that there's only one isomorphism relating the two isomorphic objects.

The terminal object is the object with one and only one morphism coming to it from any object in the category. Or alternatively, $T$ is terminal if for every object $X$ in $C$ there exists a single morphism $X \to T$.

So if $T_1$ and $T_2$ are two terminal objects, we need to prove that there is ONLY ONE isomorphism that relates $T_1$ and $T_2$.

By definition, "$T$ is terminal if for every object $X$ in $C$ there exists a single morphism $X \to T$."

Hence there should be a morphism from $T_1$ to $T_2$ and a morphism from $T_2$ to $T_1$ (apart from identity morphisms on $T_1$ and $T_2$).

There cannot be more than one morphism between $T_1$ and $T_2$.

Hence $T_1$ and $T_2$ has a unique up to a unique isomorphism.

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  • $\begingroup$ You still need to argue that $T_1 \to T_2$ and $T_2 \to T_1$ are mutual inverses. Hint: What is $Hom(T_i,T_i)$? $\endgroup$
    – Exit path
    Dec 20, 2017 at 5:38
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    $\begingroup$ There "should be" a morphism from $T_1$ to $T_2$. That's weak; there is exactly one morphism from $T_1$ to $T_2$ as $T_2$ is terminal. But to show this is an isomorphism, one needs to prove that its compositions with the unique morphism from $T_2$ to $T_1$ are identity morphisms. $\endgroup$ Dec 20, 2017 at 5:40
  • $\begingroup$ @LordSharktheUnknown Yes as T1 and T2 are terminal there is exactly one morphism: one from T1->T2 and T2->T1, Now T1 and T2 have identity functions and T1->T2, T2->T1 morphism exist, which are inverse of each other. Hence it can be proven that T1 and T2 are isomorhpic. Is this OK? $\endgroup$
    – coder_bro
    Dec 20, 2017 at 5:52
  • $\begingroup$ Are the morphisms from $T_1$ to $T_2$ and from $T_2$ to $T_1$ really inverses? $\endgroup$ Dec 20, 2017 at 5:55
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    $\begingroup$ @Ngm Exactly. So can you finish the argument now? Two objects $X$ and $Y$ are isomorphic if and only if there exist morphisms $f : X \to Y$ and $g: Y \to X$ such that $f \circ g = id_Y$ and $g \circ f= id_X$. $\endgroup$
    – Exit path
    Dec 20, 2017 at 6:12

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