#Python 2, 88 bytes

Simple solution.

s=input()
m=map(lambda x,y:y>x,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

Try it online

If same letters in a row were neither a rise nor a fall, the solution would be 79 bytes:

s=input()
m=map(cmp,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

Python 2, 88 bytes

Simple solution.

s=input()
m=map(lambda x,y:y>x,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

Try it online

If same letters in a row were neither a rise nor a fall, the solution would be 79 bytes:

s=input()
m=map(cmp,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

#Python 2, 88 bytes

Simple solution.

s=input()
m=map(lambda x,y:y>x,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

Try it online

If same letters in a row was neither rise nor fall, the solution is 79 bytes:

s=input()
m=map(cmp,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

#Python 2, 88 bytes

Simple solution.

s=input()
m=map(lambda x,y:y>x,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))

Try it online

If same letters in a row were neither a rise nor a fall, the solution would be 79 bytes:

s=input()
m=map(cmp,s[:-1],s[1:])
print all(x-y for x,y in zip(m[:-1],m[1:]))