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I seem to remember a proof that if a category $C$ has coproducts and a $0$ object, then necessarily if we had objects of $C$, say $a$ and $-a$, such that $a \oplus -a \simeq 0$, then $a\simeq 0\simeq -a$.

But right now, I can't place this, nor am I 100% sure that that is the correct property. I am able to show that, using the various universal properties at play, that the morphisms in such a category are necessarily quite boring, but not that it completely collapses (i.e. if I assume that all objects have a corresponding 'negative' object).

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With Yoneda ? For every object $X$, $Mor(a\oplus (-a),X)$ is a singleton since $a \oplus (-a)$ is initial. And $Mor(a\oplus (-a),X) \cong Mor(a,X) \times Mor((-a),X)$ is a singleton as well. So $Mor(a,X) \cong Mor((-a),X) \cong Mor(0,X)$. Hence $a \cong (-a) \cong 0$.

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  • $\begingroup$ So all I need is that $a \oplus -a$ is initial for this to work? $\endgroup$
    – Jacques Carette
    Commented Jun 20, 2012 at 17:57
  • $\begingroup$ Yes, the only hypothesis is $a \oplus (-a)$ initial. $\endgroup$
    – Philippe Gaucher
    Commented Jun 20, 2012 at 18:05
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There are three proofs:

1) If $C$ has infinite coproducts, you may use the Eilenberg swindle:

$$a \cong a \oplus (-a \oplus a) \oplus (-a \oplus a) \oplus \dotsc \cong (a \oplus -a) \oplus (a \oplus -a) \oplus \dotsc \cong 0.$$

2) Here is my personal summary of Philippe's proof: (a) It is enough to prove the dual statement $a \times a^{-1} \cong 1 \Rightarrow a \cong 1$ in categories with products. (b) This may be reduced, by the Yoneda embedding which preserves products, to the case $\mathsf{Set}$. (c) In the case of $\mathsf{Set}$ it is clear.

3) If $\mathcal{L}$ is an invertible object of a symmetric monoidal category, then $\mathcal{L} \otimes -$ is cocontinuous since $\mathcal{L}^{-1} \otimes - $ is right adjoint to it. In particular $\mathcal{L} \otimes -$ preserves initial objects. Now apply this to $(C,\oplus,0)$. Thus if $a$ is invertible w.r.t. $\oplus$, we have $a \oplus 0 \cong 0$. On the other hand, the left hand side is also $a$.

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for any object $X\in C$ we have the bijections $C(a, X)\times C(-a, X) \cong C(a\oplus -a, X)\cong C(0, X)=${$0$} then the sets $C(a, X), C(-a, X)$ have one element (the 0-morphism $a\to X=a\to 0\to X$ and $-a\to X=-a\to 0\to X$). THen $a$ is a initial object as $0$ then $a\cong 0$, similary $-a\cong 0$

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  • $\begingroup$ I posted too late the answere.. $\endgroup$
    – Buschi Sergio
    Commented Jun 20, 2012 at 18:09
  • $\begingroup$ By a few minutes, yes. $\endgroup$
    – Jacques Carette
    Commented Jun 20, 2012 at 18:11

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