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Hi I'm having problems with coming up with a proof for this simple property of cartesian closed categories (CCC) and exponential objects, namely that for any object $a$ in a CCC $C$ with an initial object $0$, $a$ is isomorphic to $a^1$ where $1$ is the terminal object of $C$. In most of the category theory books i've read this is usually left as an exercise, but for some reason I can't get a handle on it.

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  • $\begingroup$ Try considering the fact that $a\times 1\approx a$, and use it in the diagram on the definition here: en.wikipedia.org/wiki/Exponential_object $\endgroup$
    – guaraqe
    May 23, 2012 at 17:55
  • $\begingroup$ Juan's answer seems to be on the right track here (I'm using Rob Goldblatt's Topoi which doesn't mention functors or adjoints till much much later on, so the other two answers don't really help me). However I'm still not clear as to how making the function $g$ in the diagram an isomorphism gets me much further. Would it be possible to have a little more detail on the proposed solution? Apologies again if this isn't the right place to ask this. $\endgroup$
    – Anas
    May 23, 2012 at 21:56
  • $\begingroup$ @Anas: Hi, and welcome to math.SE! Users with any number of "reputation points" can comment on their own questions and answers (once you obtain 50 points, you gain the ability to comment anywhere), but you were not able to comment because you were not signed into the account that asked the question. I've now merged the duplicate account into the original. Here is an explanation of reputation points. $\endgroup$ May 23, 2012 at 22:31
  • $\begingroup$ @ZevChonoles thanks for clearing that up and merging the accounts. $\endgroup$
    – Anas
    May 23, 2012 at 22:40

3 Answers 3

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For any object $x$, we have: $$\operatorname{Hom}(x,a^1)\cong \operatorname{Hom}(x\times 1,a)\cong \operatorname{Hom}(x,a)$$ So Yoneda's lemma gives us that $a$ is isomorphic to $a^1$.

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    $\begingroup$ Thanks for the answer, but I was looking for a proof that didn't use Yoneda's lemma, or the Hom notation, since the book I'm working from hasn't introduced either of those yet. $\endgroup$
    – Anas
    May 23, 2012 at 17:47
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    $\begingroup$ @Anas: I'm stunned that there's a category theory text that gets to cartesian closed categories before hom-sets or even Yoneda. For one thing, how do you even define exponential objects? $\endgroup$
    – Zhen Lin
    May 23, 2012 at 18:40
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    $\begingroup$ @ZhenLin Yeah I'm using Robert Goldblatt's Topoi: the Categorial Analysis of Logic, and from what I understand it's a reasonably controversial book amongst category theorists because of its approach to the subject. $\endgroup$
    – Anas
    May 23, 2012 at 22:30
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You can also reason as follows, without the Yoneda lemma. But proving uniqueness of right adjoints is cumbersome without using Yoneda, and easy with. Anyway, here it goes:

The functor $(-)\times 1$ is isomorphic to the identity functor. The identity functor is a right adjoint of itself, so the identity functor is also right adjoint to $(-)\times 1$. Then uniqueness of right adjoints gives that $(-)^1$ is isomorphic to the identity functor.

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I know that this is a bit late, but I'm readong the book right now and found this post, so i'll give my solution anyway, which doesn't use the Yoneda Lemma, or adjunctions, etc.

The idea is to notice that $id_a : a\to a$ is a terminal object in the comma category $C\to a$. But the definition of the exponential allows you to show that $ev : a^1 × 1 \to a$ is also a terminal object, and therefore isomorphic to the first one. It is then easy to conclude that $a^1×1 \simeq a$ and so $a^1 \simeq a$

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