Part 1.
Part 2.
Let g, h 
H.
Since g, h G and f is a morphism
of groups, we have
f(gh) = f(g) f(h),
so the restriction of f to H is also a morphism of groups.
H.
Since g, h G and f is a morphism
of groups, we have
f(gh) = f(g) f(h),
so the restriction of f to H is also a morphism of groups.
To see that g' is a bijection for g G, let y
Y. Then there is an element x
X with g(x) = y. Since y Y and g has inverse
g-1 in G, we have
g-1(y) Y, and so
x Y. Since
g'(x) = y, this proves that g' is surjective.
(gh)'(y ) = (gh)(y ) =
g(h(y)) = g'(h'(y)) =
(g'h')(y),
G, let y
Y. Then there is an element x
X with g(x) = y. Since y Y and g has inverse
g-1 in G, we have
g-1(y) Y, and so
x Y. Since
g'(x) = y, this proves that g' is surjective.
Clearly, g' is injective because g is.
It remains
to show that the map g -> g' is a morphism. Let
h G and y Y. Then
proving (gh)' = g'h'.