**Describe how the Möbius band is related to $ℝP^2$:**
Lecture notes page 47

- Take any $x∈U$, we find a open neighbourhood of $x$ in $U$, then $U$ will be open.

$X$ is Hausdorff, take $y≠x$, there is open sets $U_x∋x,U_y∋y$, so $F=X∖U_y$ is closed, by assumption $U∩F$ is open in $F$, but $U_x⊂F$, then $U∩U_x$ is open in $U_x$.

Since $U_x$ is open, $U∩U_x$ is also open in $X$. - Let $X$ be the cofinite topology on ℕ, then $2ℕ$ is not open.

For any closed $F⊊X$, $F$ is finite.

$F∩2ℕ=F∩U$, where $U=2ℕ∪(X∖F)$ is open.

So $F∩2ℕ$ is open in $F$.

- Define a continuous bijection $f_r(θ):[0,2π)→[0,2π)$
\[f_r(θ)=\begin{cases}(1-r)θ&0≤θ≤π\\2π-(1+r)(2π-θ)&π≤θ<2π\end{cases}\]
so that $f_r([0,π])=[0,(1-r)π],f([π,2π))=[(1-r)π,2π)$

Then $(r,θ)↦(r,f_r(-θ))$ is a homeomorphism from $X_3$ to $X_1$.

- For $b=1$, the set $[0,e^{πi})$ is not open in $S^1$ but its image $[0,π)$ under $f^{-1}$ is open in $[0,1)$, so $f$ is not a quotient map.

Error! Click to view log.

For $b>1$, for any open set $U$ in $[0,b)$,

If $x=0$, take $y∈U∩(0,1)$ then $f(x)$ lies in an open set $f(U∩(y,b))$.

If $x>0$, take $y∈U∩(0,x)$ then $f(x)$ lies in an open set $f(U∩(y,b))$.

So $f(U)$ is open, $f$ is a quotient map.