In a tensor category, does $Xotimes Ycong 0$ imply $Ycong 0$ for non-zero $X$?












0












$begingroup$


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 $Xotimes Y cong 0$ 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 $otimes$






category-theory abelian-categories monoidal-categories functors







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 8:14






  • 1




    $begingroup$
    $mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:27










  • $begingroup$
    @QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:51






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 9:04










  • $begingroup$
    @QiaochuYuan: Vect x Vect does not have End(1) = k!
    $endgroup$
    – Noah Snyder
    Jan 11 at 18:30
















0












$begingroup$


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 $Xotimes Y cong 0$ 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 $otimes$






category-theory abelian-categories monoidal-categories functors







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 8:14






  • 1




    $begingroup$
    $mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:27










  • $begingroup$
    @QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:51






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 9:04










  • $begingroup$
    @QiaochuYuan: Vect x Vect does not have End(1) = k!
    $endgroup$
    – Noah Snyder
    Jan 11 at 18:30














0












0








0





$begingroup$


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 $Xotimes Y cong 0$ 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 $otimes$






category-theory abelian-categories monoidal-categories functors







share|cite|improve this question











$endgroup$




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 $Xotimes Y cong 0$ 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 $otimes$






category-theory abelian-categories monoidal-categories functors




category-theory abelian-categories monoidal-categories functors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 26 '18 at 8:10







Jo Be

















asked Jul 26 '18 at 7:54









Jo BeJo Be

6751519




6751519








  • 2




    $begingroup$
    No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 8:14






  • 1




    $begingroup$
    $mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:27










  • $begingroup$
    @QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:51






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 9:04










  • $begingroup$
    @QiaochuYuan: Vect x Vect does not have End(1) = k!
    $endgroup$
    – Noah Snyder
    Jan 11 at 18:30














  • 2




    $begingroup$
    No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 8:14






  • 1




    $begingroup$
    $mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:27










  • $begingroup$
    @QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
    $endgroup$
    – Jo Be
    Jul 26 '18 at 8:51






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Qiaochu Yuan
    Jul 26 '18 at 9:04










  • $begingroup$
    @QiaochuYuan: Vect x Vect does not have End(1) = k!
    $endgroup$
    – Noah Snyder
    Jan 11 at 18:30








2




2




$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14




$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14




1




1




$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27




$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27












$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51




$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51




1




1




$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04




$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04












$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30




$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30










1 Answer
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$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
    $endgroup$
    – Jo Be
    Jan 11 at 18:47












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
    $endgroup$
    – Jo Be
    Jan 11 at 18:47
















1












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
    $endgroup$
    – Jo Be
    Jan 11 at 18:47














1












1








1





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 18:41









Noah SnyderNoah Snyder

7,65232955




7,65232955












  • $begingroup$
    Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
    $endgroup$
    – Jo Be
    Jan 11 at 18:47


















  • $begingroup$
    Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
    $endgroup$
    – Jo Be
    Jan 11 at 18:47
















$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47




$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47


















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