PMath 336: Introduction to Group Theory
Exercise set 1
May 13, 2008
Solution should be submitted by the end of the Monday lecture of the following week, either in class or
in the submission box. You may not submit joint or identical works.
1. (6 points) Let
f
:
X
→
Y
be a function between two sets. Show that
f
is injective if and only if, for
any set
Z
, and any two functions
h, g
:
Z
→
X
, if
f
◦
h
=
f
◦
g
then
h
=
g
(in other words,
f
is injective
iff it can be cancelled on the left.) Likewise, show that
f
is surjective if and only if, for any set
Z
and
functions
h, g
:
Y
→
Z
, if
h
◦
f
=
g
◦
f
, then
h
=
g
(
f
can be cancelled on the right.)
Solution:
We showed in class that
f
is injective if and only if there is a function
t
:
Y
→
X
with
t
◦
f
the identity (if
X
is empty, the statement is easy.) Applying
t
to the equality
fh
=
fg
we get
h
=
g
. Conversely, if
f
is not injective,
f
(
x
) =
f
(
y
) for some distinct
x
and
y
. Consider
Z
=
{
1
}
,
and
h, g
the functions mapping 1 to
x, y
, respectively.
If
f
is surjective, let
h
and
g
be two functions from
Y
,
y
an element of
Y
. Then
y
=
f
(
x
) for some
x
∈
X
. Hence
h
(
y
) =
h
(
f
(
x
)) =
g
(
f
(
x
)) =
g
(
y
), so
h
and
g
agree on all elements, hence they are
equal. Conversely, if
y
∈
Y
is not in the image of
f
, let
Z
=
{
0
,
1
}
,
h
:
Y
→
Z
the constant function
0, and
g
the function such that
g
(
y
) = 1 is 0 for all other values. Then
hf
=
gf
= 0, but
h
and
g
are different.
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 Spring '08
 MOSHEKAMENSKY
 Group Theory, Inverse element

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