Question 32850
 You really did not know how to type --> , did you ???

a ) f: Z --> Z defined by f(x)= -x
b) f: Z2 -->Z2, defined by f(x)= -x

 Clearly a,b) are additive homo. (too simple)

 
c) g: Q --> Q defined by g(x) = (1/(x^2+1))
  since g(1) = 1/2 ~= 1 , so g cannot be a multiplicative homo
 [ Note g(1) must be equal to 1 ] 
 
d) h: R --> M2x2(R), defined by h(a)= {(-a,a) , (0,0)}
  check h(a+b) = ( -(a+b) , (a+b))
                 (0 ,      0)
    = (-a , a)  + (-b, b)
      (0  , 0)    (0, 0)
   = h(a) + h(b)
  So, h is a homo from (R, +) to (M2x2(R), +)
   
e) f: Z12 --> Z4, defined by f([x]12) = [x]4 
  First of all, you have to know that f is weel-defined, i.e.
 [x]1= [y]12 --> x = y mod 12 --> x = y mod 4 --> [x]4 = [y]4
 
  f([x]12+ [y]12) = f([x+y}12) = [x+y]4 =  [x}4+ [y]4 = f([x]12)+ f([y]12)

 So, f is a homo. from Z12 to Z4.

 Of course, you have to work hard.

 Kenny