❖
✖ Questions

- Howto

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★ | right first attempt |

✓ | right |

✗ | wrong |

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 $q:X\to Y$
from a topological space $X$
to $Y$
equipped with the quotient topology. Which statements are true?
*There is at least one mistake.*

For example, choice (a) should be False.
*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 True.
*There is at least one mistake.*

For example, choice (d) should be True.
*There is at least one mistake.*

For example, choice (e) should be True.
*There is at least one mistake.*

For example, choice (f) should be False.
*Correct!*

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be True.

For example, choice (d) should be True.

For example, choice (e) should be True.

For example, choice (f) should be False.

Consider
$q:\left\{-1,1\right\}\to \left\{0\right\}$ the constant function.
Then for every $f:Z\to \left\{-1,1\right\}$
the composition $q\circ f$
is constant as well, hence continuous. But, of course, we may choose
$f$ to be
non-continuous, e.g. $f=sign:\mathbb{R}\to \left\{-1,1\right\}$.

*False*$q$ is usually not bijective. Hence, an inverse does not necessarily exist.*False*$q$ 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*$q$ is continuous by the definition of the quotient topology. Hence, the composition $q\circ f$ is continuous as well.*False*Consider $q:\left\{-1,1\right\}\to \left\{0\right\}$ the constant function. Then for every $f:Z\to \left\{-1,1\right\}$ the composition $q\circ f$ is constant as well, hence continuous. But, of course, we may choose $f$ to be non-continuous, e.g. $f=sign:\mathbb{R}\to \left\{-1,1\right\}$.

Which of the following statements are true?
*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 True.
*There is at least one mistake.*

For example, choice (c) should be False.
*Correct!*

For example, choice (a) should be True.

For example, choice (b) should be True.

For example, choice (c) should be False.

Consider any infite
$X$
with the cofinite topology and an infinite subset
$A\subset X$. Then the quotient
topology on $X\u2215A$ is not the
cofinite topology. Indeed, $V=X\u2215A\setminus \left\{A\right\}\subset X\u2215A$
is open with respect to the cofinite topology on
$X\u2215A$,
but not open with respect to the quotient topology, as
${q}^{-1}\left(V\right)=X\setminus A$ has an infinite
complement, namely $A$.

*True*Assume $q:X\to Y$ is the quotient map. Since every subset $U\subset X$ is open (since $X$ is discrete) also the preimage ${q}^{-1}\left(V\right)$ of every subset $V\subset Y$ is open. Hence, by definition every subset $V\subset Y$ is open.*True*Note, that $q:X\to Y$ was assumed to be surjective. Hence, ${q}^{-1}\left(V\right)=\varnothing $ only for $V=\varnothing $ and ${q}^{-1}\left(V\right)=X$ only for $V=Y$. Hence, only $Y$ and $\varnothing $ are open.*False*Consider any infite $X$ with the cofinite topology and an infinite subset $A\subset X$. Then the quotient topology on $X\u2215A$ is not the cofinite topology. Indeed, $V=X\u2215A\setminus \left\{A\right\}\subset X\u2215A$ is open with respect to the cofinite topology on $X\u2215A$, but not open with respect to the quotient topology, as ${q}^{-1}\left(V\right)=X\setminus A$ has an infinite complement, namely $A$.

Consider the topological space $\left(X,\tau \right)$
with $X=\left\{a,b,c,d,e\right\}$ and
$\tau =\left\{X,\varnothing ,\left\{a\right\},\left\{a,b\right\},\left\{c,d,e\right\}\right\}$ and the
equivalence relation $\sim $
induced by $a\sim b$ and
$c\sim d$. Which subsets of
$X\u2215\sim \phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\left\{\left[a\right],\left[c\right],\left[e\right]\right\}$ 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!*

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be False.

For example, choice (f) should be True.

${q}^{-1}\left(\left\{\left[c\right],\left[e\right]\right\}\right)={q}^{-1}\left(\left\{\left\{c,d\right\},\left\{e\right\}\right\}\right)=\left\{c,d,e\right\}\in \tau $

*True*${q}^{-1}\left(\left\{\left[a\right]\right\}\right)={q}^{-1}\left(\left\{\left\{a,b\right\}\right\}\right)=\left\{a,b\right\}\in \tau $*False*${q}^{-1}\left(\left\{\left[c\right]\right\}\right)={q}^{-1}\left(\left\{\left\{c,d\right\}\right\}\right)=\left\{c,d\right\}\notin \tau $*False*${q}^{-1}\left(\left\{\left[e\right]\right\}\right)={q}^{-1}\left(\left\{\left\{e\right\}\right\}\right)=\left\{e\right\}\notin \tau $*False*${q}^{-1}\left(\left\{\left[a\right],\left[c\right]\right\}\right)={q}^{-1}\left(\left\{\left\{a,b\right\},\left\{c,d\right\}\right\}\right)=\left\{a,b,c,d\right\}\notin \tau $*False*${q}^{-1}\left(\left\{\left[a\right],\left[e\right]\right\}\right)={q}^{-1}\left(\left\{\left\{a,b\right\},\left\{e\right\}\right\}\right)=\left\{a,b,e\right\}\notin \tau $*True*${q}^{-1}\left(\left\{\left[c\right],\left[e\right]\right\}\right)={q}^{-1}\left(\left\{\left\{c,d\right\},\left\{e\right\}\right\}\right)=\left\{c,d,e\right\}\in \tau $

On the disc ${D}^{2}$ we define an
equivalence relation $\sim $ by
$x\sim y\iff \left|x\right|=\left|y\right|$. Then the identification
space ${D}^{2}\u2215\sim $ is homeomorphic
to

On the disc ${D}^{2}$ we define
an equivalence relation $\sim $
by $x\sim y\iff x=y\text{or}\left|x\right|=\left|y\right|=1.$ Then the
identification space ${D}^{2}\u2215\sim $
is homeomorphic to
*Choice (a) is Incorrect.*
*Choice (b) is Incorrect.*
*Choice (c) is Incorrect.*
*Choice (d) is Correct!*
*Choice (e) is Incorrect.*

On the disc ${D}^{2}$ we define
an equivalence relation $\sim $
by $x\sim y\iff x=y\text{or}\left(\left|x\right|\le \frac{1}{2}\text{and}\left|y\right|\le \frac{1}{2}\right).$ Then the
identification space ${D}^{2}\u2215\sim $
is homeomorphic to
*Choice (a) is Incorrect.*
*Choice (b) is Correct!*
*Choice (c) is Incorrect.*
*Choice (d) is Incorrect.*
*Choice (e) is Incorrect.*

A homeomorphism
is induced by ${D}^{2}\to {D}^{2},x\mapsto \left\{\begin{array}{cc}0\phantom{\rule{1em}{0ex}}\hfill & \left|x\right|\le \frac{1}{2}\hfill \\ \frac{2\left(\left|x\right|-1\u22152\right)}{\left|x\right|}\cdot x\phantom{\rule{1em}{0ex}}\hfill & \left|x\right|\ge \frac{1}{2}.\hfill \end{array}\right.$

On the disc ${D}^{2}$ we define
an equivalence relation $\sim $
by $\left({x}_{1},{x}_{2}\right)\sim \left({y}_{1},{y}_{2}\right)\iff {x}_{2}={y}_{2}\text{or}\left|{x}_{2}\right|=\left|{y}_{2}\right|=1.$ Then the
identification space ${D}^{2}\u2215\sim $
is homeomorphic to
*Choice (a) is Incorrect.*
*Choice (b) is Incorrect.*
*Choice (c) is Correct!*
*Choice (d) is Incorrect.*
*Choice (e) is Incorrect.*