Proof
(1). For
,
we have
-
(2) follows directly from
.
(3). Let
and
-
Because of the surjectivity of , there exist for every
a
such that
.
Therefore
-
and is itself the zero mapping. Due to
fact,
injective.
(4). The condition means that we may consider
as a
linear subspace.
Because of
fact,
we can write
-
with another -linear subspace
.
A linear form
-
can always be extended to a linear form
-
for example, by defining on to be the zero form. This means the surjectivity.