Introduction to Advanced Mathematics — Midterm II
November 3rd, 2011
Write your answers in the space provided after each question. Please make an effort to write neatly.
The problems marked by
◦
refer directly to material in the class notes or your homework. You
may want to start with those questions—they make up more than half of the test.
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KEY
1
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◦
Let
f
:
A
→
B
be a function. Give the deﬁnition of
leftinverse
of
f
, and the deﬁnition of
rightinverse
of
f
.
–A function
g
:
B
→
A
is a
leftinverse
of
f
if
g
◦
f
= 1
A
.
–A function
g
:
B
→
A
is a
rightinverse
of
f
if
f
◦
g
= 1
B
.
(Class notes, Deﬁnition 12.1.)
◦
Let
f
:
A
→
B
be a function. Prove that if
f
has a rightinverse, then
f
is surjective. (We
have also proved that if
f
is surjective, then it has a rightinverse. You don’t need to prove
this now.)
Let
g
:
B
→
A
be a rightinverse of
f
: therefore,
f
◦
g
= 1
B
. Let
b
∈
B
be any element. We
have to show that there is an element
a
∈
A
such that
f
(
a
) =
b
. I claim that
g
(
b
)
is such an
element. Indeed,
f
(
a
) =
f
(
g
(
b
)) =
f
◦
g
(
b
) = 1
B
(
b
) =
b
as needed.
(This is the argument given in the proof of Lemma 13.2, for the particular case
g
◦
f
= 1
A
.)
•
Let
f
:
A
→
B
be a function. Prove that
f
is injective if and only if for all subsets
S
⊆
A
,
S
=
f

1
(
f
(
S
))
.
First assume that
S
=
f

1
(
f
(
S
))
for all subsets
S
⊆
A
. In order to prove that
f
is injective,
we have to show that if
f
(
a
) =
f
(
a
0
)
, then
a
=
a
0
. Let
S
=
{
a
}
. Since
f
(
a
0
) =
f
(
a
)
, we
have
a
0
∈
f

1
(
f
(
S
))
. Since
f

1
(
f
(
S
)) =
S
=
{
a
}
, this says that
a
0
∈ {
a
}
, and it follows
that
a
0
=
a
, as needed.
Next, assume that
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 Fall '11
 Aluffi
 Inverse function, Det

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