# In a tensor category, does $X\otimes Y\cong 0$ imply $Y\cong 0$ for non-zero $X$?

By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $\operatorname{Hom}(1,1)\cong k$.$^1$

Suppose we fix some non-zero object $X$ in such a category, and we take any old object $Y$. Can we conclude from $X\otimes Y \cong 0$ that $Y$ has to be zero?

I know that in this setting the tensor product is (bi)exact. And I think the statement would be obvious if $\otimes$ would reflect isomorphisms, but I don't feel it does.

Any hints?

$^1$ This question really only needs rigid abelian with bilinear $\otimes$

• No, this is already false in $\text{Vect} \times \text{Vect}$ with the pointwise tensor product. – Qiaochu Yuan Jul 26 '18 at 8:14
• $\mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)\otimes (W_1,W_2) \equiv (V_1\otimes W_1, V_2\otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)\otimes (0,W)\cong (0,0)$, right? – Jo Be Jul 26 '18 at 8:27
• @QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind? – Jo Be Jul 26 '18 at 8:51
• Yes, that's right. – Qiaochu Yuan Jul 26 '18 at 9:04
• @QiaochuYuan: Vect x Vect does not have End(1) = k! – Noah Snyder Jan 11 at 18:30

The coevaluation map $$1 \rightarrow {}^*X \otimes X$$ is non-zero, so by simplicity of $$1$$ it's injective. But then by biexactness we have: $$Y \hookrightarrow {}^* X \otimes X \otimes Y \cong 0.$$