My problem is, that my mathematical understanding is full of gaps and I am stuck at some steps or I dont know if I am finished or not.

Sorry for the text.

a)f:ℝ³-->ℝ² f((x,y,z)) = (xy,x+y)

f(0)ofℝ³ = f(0,0,0) = (0*0, 0+0) = (0,0) = ℝ²

Showing homomorphism

v1=(x,y,z), v2=(x',y',z') ∈ℝ³Left side:

f((x,y,z)+(x',y',z')) = f(x+x', y+y', z+z') = ((x+x')*(y+y'), (x+x')+(y+y'))

Right side:

f(x,y,z)+f(x',y',z') = (xy, x+y) + (x'y', x'+y') = ((xy+x'y'), (x+y+x'+y'))

Am I done here or can I transform the term to make both parts equal?

b)f:ℝ³-->ℝ³ f((x,y,z)) = (0, -x+y, 3x-5y)

f(0)ofℝ³ = f(0,0,0) = (0, 0+0, 3*0-3*0) = (0,0,0) = ℝ³

Showing homomorphism

v1=(x,y,z), v2=(x',y',z') ∈ℝ³

Left side:

f((x,y,z)+(x',y',z')) = f(x+x', y+y', z+z') = (0 , ((-x+(-x'))+(y+y')) , (3*(x+x')-5*(y+y')))

Right side:

f(x,y,z)+f(x',y',z') = (0 , -x+y , 3x-5y) + (0 , -x'+y' , 3x'-5y') = (0 , ((-x+(-x'))+(y+y')) , (3*(x+x')-5*(y+y')))

Its the same as far as I see it. It this correct and can I continue the proof now?

c)f:ℝ³-->ℝ³ f((x,y,z)) = (0,0,1)

f(0)ofℝ³ = f(0,0,0) ≠ (0,0,1)

- Not a linear function between ℝ³ and ℝ³