Can we extend linearly for the twist map ? Is this a linear map?
Yes, it’s supposed to be a linear map (otherwise its definition would be incomplete).
Darij is right; I should have said that the twist map is meant to be linear.
In fact (as is often done), I am being too brief in the definition of . Remember that if I want to define a linear map to a vector space , I cannot define arbitrarily on the pure tensors . What I really need to do is define a bilinear map such that “descends” to . I *can* define by defining arbitrarily for bases and of and respectively, and then extending bilinearly.
I am not asking you to do this explicitly in this problem, but you should make sure that you know how to do it. (The first time you do this, it may seem a bit tricky. Once you do this kind of thing enough times, it achieves the status of “obvious”, and is therefore omitted.)
Federico, is it sufficient to use commuting diagrams to prove 2bi, 2bii and 2biii as you did in your lecture?
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