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Mathematics
Linear and Abstract Algebra
Tensor Products of Modules - Bland - Remark, Page 65
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[QUOTE="andrewkirk, post: 5491335, member: 265790"] Good morning MA. I wonder if you are affected by the wild weather this morning from the East Coast Low in the Tasman Sea. In the example given, the purported definition of [itex]g[/itex] is that [itex]g(x\otimes y)=f(x)\otimes y[/itex]. If we think about what this means, it is saying that $$g\Big(\chi_{(x,y)}+K\Big)=\chi_{(f(x),y)}+K'$$ where [itex]K[/itex] and [itex]K'[/itex] are the submodules of [itex]Z[/itex] and [itex]Z'[/itex] respectively that have the required 'zero' properties inherited from the modules [itex]M,M',N[/itex]; and [itex]Z[/itex] and [itex]Z'[/itex] are the free R-modules generated by [itex]{\{\chi_{(u,v)}\ :\ u\in M\wedge v\in N\}}[/itex] and [itex]{\{\chi_{(u,v)}\ :\ u\in M'\wedge v\in N\}}[/itex] respectively. For this to be well-defined, we require that, for any [itex]a,c\in M[/itex] and [itex]b,d\in N[/itex] such that [itex](a,b)[/itex] and [itex](c,d)[/itex] are in the same coset of [itex]K[/itex], it will be the case that [itex](f(a),b))[/itex] and [itex](f(c),d)[/itex] are in the same coset of [itex]K'[/itex]. Otherwise, [itex]g[/itex] will not be a function because [itex]g\Big(\chi_{(a,b)}+K\Big)\neq g\Big(\chi_{(c,d)}+K\Big)[/itex] even though [itex]\chi_{(a,b)}+K=\chi_{(c,d)}+K[/itex]. Hence we have two different output values for a single input. The argument involving the diagram appears to be a proof that [itex]g[/itex], thus defined, is indeed well-defined. [/QUOTE]
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Mathematics
Linear and Abstract Algebra
Tensor Products of Modules - Bland - Remark, Page 65
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