There is not limit to the number of attempts
for each question. However, in order to make the best use of this quiz, you
should give youself sufficient time to think about every question carefully
before submitting an answer. You should not only guess the answer, but
also think about an argument that supports your choice. You will then be
able to compare your justification with the argument in the provided model
solution.
For some of the questions credit goes to Michael Eisermann.
Start Quiz
Consider a surjective map
from a topological space
to
equipped with the quotient topology. Which statements are true?
There is at least one mistake.For example, choice (a) should be
False.
is usually not bijective. Hence, an inverse does not necessarily exist.
There is at least one mistake.For example, choice (b) should be
False.
is usually not bijective. Hence, an inverse does not necessarily exist.
There is at least one mistake.For example, choice (c) should be
True.
This is one direction of the universal property.
There is at least one mistake.For example, choice (d) should be
True.
This is the other direction of the universal property.
There is at least one mistake.For example, choice (e) should be
True.
is
continuous by the definition of the quotient topology. Hence, the composition
is continuous
as well.
There is at least one mistake.For example, choice (f) should be
False.
Consider
the constant function.
Then for every
the composition
is constant as well, hence continuous. But, of course, we may choose
to be
non-continuous, e.g. .
Correct!
- False
is usually not bijective. Hence, an inverse does not necessarily exist.
- False
is usually not bijective. Hence, an inverse does not necessarily exist.
- True This is one direction of the universal property.
- True This is the other direction of the universal property.
- True is
continuous by the definition of the quotient topology. Hence, the composition
is continuous
as well.
- False Consider
the constant function.
Then for every
the composition
is constant as well, hence continuous. But, of course, we may choose
to be
non-continuous, e.g. .
Which of the following statements are true?
There is at least one mistake.For example, choice (a) should be
True.
Assume
is the quotient map.
Since every subset
is open
(since
is discrete) also
the preimage
of every
subset
is open. Hence, by
definition every subset
is open.
There is at least one mistake.For example, choice (b) should be
True.
Note, that
was assumed to be
surjective. Hence,
only for
and
only
for
. Hence,
only
and
are open.
There is at least one mistake.For example, choice (c) should be
False.
Consider any infite
with the cofinite topology and an infinite subset
. Then the quotient
topology on is not the
cofinite topology. Indeed,
is open with respect to the cofinite topology on
,
but not open with respect to the quotient topology, as
has an infinite
complement, namely .
Correct!
- True Assume
is the quotient map.
Since every subset is open
(since is discrete) also
the preimage of every
subset is open. Hence, by
definition every subset
is open.
- True Note, that
was assumed to be
surjective. Hence,
only for
and only
for . Hence,
only
and
are open.
- False Consider any infite
with the cofinite topology and an infinite subset
. Then the quotient
topology on is not the
cofinite topology. Indeed,
is open with respect to the cofinite topology on
,
but not open with respect to the quotient topology, as
has an infinite
complement, namely .
Consider the topological space
with
and
and the
equivalence relation
induced by
and
. Which subsets of
are open with respect to
the quotient topology?
There is at least one mistake.For example, choice (a) should be
True.
There is at least one mistake.For example, choice (b) should be
False.
There is at least one mistake.For example, choice (c) should be
False.
There is at least one mistake.For example, choice (d) should be
False.
There is at least one mistake.For example, choice (e) should be
False.
There is at least one mistake.For example, choice (f) should be
True.
Correct!
- True
- False
- False
- False
- False
- True
On the disc
we define an
equivalence relation
by
. Then the identification
space
is homeomorphic
to
Choice (a) is Correct!
The homeomorphism
is induced by
.
On the disc
we define
an equivalence relation
by
Then the
identification space
is homeomorphic to
Choice (d) is Correct!
Indeed, the homeomorphisms is given in the lecture notes. It is induced by
On the disc
we define
an equivalence relation
by
Then the
identification space
is homeomorphic to
Choice (b) is Correct!
A homeomorphism
is induced by
On the disc
we define
an equivalence relation
by
Then the
identification space
is homeomorphic to
Choice (c) is Correct!
A homeomorphism
is given by