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

Question

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?

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

Answer 1

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: $$Y \hookrightarrow {}^* X \otimes X \otimes Y \cong 0.$$