Algebra (647), Homework 1, solutions.
1. I.7: 1.
hom((X, x0),(Y, y0))
is a subset of
hom(X, Y )
in the category of sets. So
in particular it is a set.
In addition, the identity is a perfectly good map of pointed sets since
1X(x0) = x0
.
We just really need to check if
f: (X, x0)→(Y, y0)
and
g: (Y, y0)→
(Z, z0)
is a map of pointed sets, then so is
g◦f
. But this is clear since
(g◦f)(x0) = g(f(x0)) = g(y0) = z0.
2. I.7: 2.
Suppose
g0
also satise
f◦g0= 1
and
g0◦f= 1
. We get
g=g◦1 = g◦(f◦g0) = (g◦f)◦g0= 1 ◦g0=g0.
It is worth observing that we just proved something stronger than what was
required. Suppose
f
is a morphism with a
left inverse g
(meaning
g◦f= 1
)
and a right inverse
g0
(meaning
f◦g0= 1
). Then
g=g0
and is a proper
(two-sided) inverse.
3. I.7: 7.
Let
A
in
C
be an object whose underlying set has more than one element.
Let
X
be a set with more than one element. Let
x1, x2
be distinct ele-
ments of
X
. Pick two distinct elements of
a1, a2∈A
and let
j:X→A
be
a set map so that
j(x1) = a1
and
j(x2) = a2
If
i:X→F
is a free object on
X
, there is a unique map
f:F→A
so
that
f◦i=j
. Since
j(x1)6=j(x2)
, it follows that
i(x1)6=i(x2)
.
So
i
sends distinct elements of
X
to distinct elements of
F
.
4. Give an example of a concrete category and a morphism f:A→Bin that
category that is an isomorphism on the underlying set, but not an isomor-
phism in the category.
Here is a very simple example. Take the category to be partially ordered
sets. Take a partially ordered set with two elements
X={a, b}
and the
only relation being
a≤a
and
b≤b
. This satises the axioms for a partial
order, though it isn't very interesting.
Now take
Y={x, y}
with
x≤y
(and of course
x≤x
,
y≤y
). Take
f:X→Y
so that
f(a) = x
and
f(b) = y
.
f
is a set bijection, and is a map
of partially ordered sets.
But
f−1
, which exists as a map of
sets
, is not a map of partially ordered
sets. So
f
has no inverse in the category of P.O. sets.
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