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).


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

  • $\begingroup$ So all I need is that $a \oplus -a$ is initial for this to work? $\endgroup$ – Jacques Carette Jun 20 '12 at 17:57
  • $\begingroup$ Yes, the only hypothesis is $a \oplus (-a)$ initial. $\endgroup$ – Philippe Gaucher Jun 20 '12 at 18:05

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 $a$.


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 $-a\cong 0$

  • $\begingroup$ I posted too late the answere.. $\endgroup$ – Buschi Sergio Jun 20 '12 at 18:09
  • $\begingroup$ By a few minutes, yes. $\endgroup$ – Jacques Carette Jun 20 '12 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.