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1. • CommentRowNumber1.
   • CommentAuthorsgarush
   • CommentTimeApr 10th 2018
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   Author: sgarush
   Format: MarkdownItexIn the article [[exterior+algebra]], in sect. 2, subsection In General, it says that the nth exterior power of an object $V$ is the cokernel of the antisymmetrization operator $$ P_{A} = \frac{1}{n!} \sum_{\sigma \in S_{n}} \mathrm{sgn}(\sigma) \sigma. $$ Shouldn't the exterior algebra be the image of this operator? If so, I am happy to edit the article accordingly.

   In the article exterior+algebra, in sect. 2, subsection In General, it says that the nth exterior power of an object $V$ is the cokernel of the antisymmetrization operator

   $$P_{A} = \frac{1}{n!} \sum_{\sigma \in S_{n}} \mathrm{sgn}(\sigma) \sigma.$$

   Shouldn’t the exterior algebra be the image of this operator? If so, I am happy to edit the article accordingly.

2. • CommentRowNumber2.
   • CommentAuthorsgarush
   • CommentTimeApr 10th 2018
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   Author: sgarush
   Format: MarkdownItexIt looks like in [[schur+functor]] what is used is the cokernel of $1 - P_A$ -- is this correct?

   It looks like in schur+functor what is used is the cokernel of $1 - P_A$ – is this correct?

3. • CommentRowNumber3.
   • CommentAuthorOscar_Cunningham
   • CommentTimeApr 10th 2018
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   Author: Oscar_Cunningham
   Format: MarkdownItexYes you're correct. At least in the context of vector spaces over a field of characteristic $0$, $\Lambda^{n}V$ is the image of $P_A$, the coimage of $P_A$, the kernel of $1-P_A$ and the cokernel of $1-P_A$. So the statement at [[Schur functor]] is correct.

   Yes you’re correct. At least in the context of vector spaces over a field of characteristic $0$, $\Lambda^{n}V$ is the image of $P_A$, the coimage of $P_A$, the kernel of $1-P_A$ and the cokernel of $1-P_A$. So the statement at Schur functor is correct.

4. • CommentRowNumber4.
   • CommentAuthorsgarush
   • CommentTimeApr 10th 2018
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   Author: sgarush
   Format: MarkdownItexGreat, thanks. I guess I'll edit the article to bring it in line with [[Schur functor]].

   Great, thanks. I guess I’ll edit the article to bring it in line with Schur functor.