Review section done

content/Math4201/Math4201_L11.md

@@ -6,7 +6,7 @@
 >
 > A: No. Consider $X=[0,2\pi)$ and $Y=\mathbb{S}^1$ with standard topology in $\mathbb{R}^2$.
 >
-> Let $f\coloneq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ($\forall f^{-1}(V)$ is open in $X$)
+> Let $f\coloneqq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ($\forall f^{-1}(V)$ is open in $X$)
 >
 > But $f^{-1}$ is not continuous, consider the open set in $X, U=[0,\pi)$. Then $f^{-1}(U)=[0,\pi)$ is not open in $Y$.