Talk:K-group, first (Ex)
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Revision as of 12:43, 31 August 2013 by Markus Land (Talk | contribs)
First we show that the determinant induces a well-defined morphism
For this we recall that is the abelianization of the infinite general linear group over . Surely, since the ring is commutative, for every we have the determinat as a map and the diagram
commutes, where is the canonical inclusion as block matrix in the upper left corner.
Hence we get a well-defined map on the colimit by standard properties of the determinant we see that this map factors over the abelianization to obtain the desired map
The fact that this map is surjective follows from the easy observation that we have the canonical map
which obviously splits the determinant.