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In a tensor category, does $Xotimes Ycong 0$ imply $Ycong 0$ for non-zero $X$?
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By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatornameHom(1,1)cong k$.$^1$
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
asked Jul 26 at 7:54
Jo Be
6531519
add a comment |Â
up vote
0
down vote
favorite
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatornameHom(1,1)cong k$.$^1$
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
asked Jul 26 at 7:54
Jo Be
6531519
2
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
1
$mathrmVect$ 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?
– Jo Be
Jul 26 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 at 8:51
1
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatornameHom(1,1)cong k$.$^1$
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
asked Jul 26 at 7:54
Jo Be
6531519
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatornameHom(1,1)cong k$.$^1$
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
asked Jul 26 at 7:54
Jo Be
6531519
edited Jul 26 at 8:10
asked Jul 26 at 7:54
Jo Be
6531519
asked Jul 26 at 7:54
Jo Be
6531519
asked Jul 26 at 7:54
Jo Be
6531519
6531519
2
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
1
$mathrmVect$ 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?
– Jo Be
Jul 26 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 at 8:51
1
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04
add a comment |Â
2
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
1
$mathrmVect$ 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?
– Jo Be
Jul 26 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 at 8:51
1
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04
2
2
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
1
1
$mathrmVect$ 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?
– Jo Be
Jul 26 at 8:27
$mathrmVect$ 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?
– Jo Be
Jul 26 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 at 8:51
@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 at 8:51
1
1
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04
add a comment |Â
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2
No, this is already false in $textVect times textVect$ with the pointwise tensor product.
– Qiaochu Yuan
Jul 26 at 8:14
1
$mathrmVect$ 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?
– Jo Be
Jul 26 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 at 8:51
1
Yes, that's right.
– Qiaochu Yuan
Jul 26 at 9:04