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Tensor product of modules 
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#19
May1012, 03:27 PM

Mentor
P: 18,040

OK, so then we wish to prove that
[tex]Ker(f_3\circ \psi \circ f_2^{1})=Im(f_2\circ \varphi \circ f_1^{1})[/tex] I'll do one side of the inclusion: Take x in the kernel, then [itex]f_3(\psi(f_2^{1}(x)))=0[/itex]. Since [itex]f_3[/itex] is an isomorphism, its kernel is 0, thus [itex]\psi(f_2^{1}(x))=0[/itex]. Thus [itex]f_2^{1}(x)\in Ker(\psi)[/itex]. By exactness, we have that [itex]Ker(\psi)=Im(\varphi)[/itex], thus there exists a [itex]y\in A[/itex] such that [itex]\varphi(y)=f_2^{1}(x)[/itex]. In other words, [itex]f_2(\varphi(y))=x[/itex]. Write [itex]z=f_1(y)[/itex], then [itex]f^{1}_1(z)=y[/itex]. Thus [tex]f_2(\varphi(f_1^{1}(z)))=f_2(\varphi(y))=x[/tex] The lefthand side is certainly in the image, thus [itex]x\in Im(f_2\circ \varphi\circ f_1^{1})[/itex]. Can you do the other inclusion?? 


#20
May1012, 03:55 PM

P: 101

Take x in [itex]Im(f_2\circ \varphi \circ f_1^{1})[/itex]. Because it's in image, then there exists y in the domain of [itex]f_2\circ \varphi \circ f_1^{1}[/itex], i.e. in A', such that [itex]f_2\circ \varphi \circ f_1^{1}(y)=x[/itex] Since [itex]f_2[/itex] is an isomorphism it's invertible, and we'll have: [itex]f_2^{1}(x)=\varphi \circ f_1^{1}(y)[/itex] Now let's apply [itex]\psi[/itex] to both sides, we'll obtain: [itex]\psi \circ f_2^{1}(x) = \psi \circ \varphi \circ f_1^{1}(y) = 0 [/itex] Since [itex] \psi \circ \varphi = o [/itex], because anything that we put into [itex]\varphi[/itex] goes to the kernel of [itex]\psi[/itex] and therefore it becomes zero! We're done! Since [itex]f_3[/itex] is a homomorphism, it sends 0 to 0, hence: [itex] f_3 \circ \psi \circ f_2^{1} (x) = f_3(0) = 0[/itex] and that says [itex] x \in Ker(f_3\circ \psi \circ f_2^{1})[/itex] It's a very neat theorem, I think it'll make a lot of things easier in future for me, and the good thing is that it works on almost all algebraic structures that I know now, because nowhere in the proof we assumed anything else except the definition of an exact sequence, kernel and image of homomorphisms. All of them stay intact in groups, rings and modules. Thanks for your help micromass. 


#21
May1012, 08:53 PM

P: 528




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