"\\phi: R^4\\longrightarrow R^2" is given by "(a,b,c,d)\\mapsto (a,b)" This map is trivially homomorphismThis map is clearly surjective and kernel is given by "{a=b=0}." So by homorphism theorem "R^4 \/Ker \\phi\\cong R^2." Now we need to show "Ker \\phi\\cong R^2." This is given by the map "(0,0,x,y)\\mapsto (x,y)" . This map is clearly surjective and its kernel is zero and hence bijective. Homomorphism is trivial.
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