Verify that $\phi(t)=e^{it}$ is a homeomorphism between $(0,2\pi)$ and $\mathbb{T} \backslash \{1\}$, where $\mathbb{T}$ denotes the unit circle in the complex plane. Conclude that unit circle minus a point is homeomorphic to the real line.

Show that $X=A(0,r,R)$ (open annulus) and $Y= \mathbb{T}$(unit circle) are not homeomorphic

Let $X=S^n \backslash \{(0,\ldots ,1)\} \subseteq \mathbb{R^{n+1}}$ and $Y= \mathbb{R^n}$ , and define

$f:X \rightarrow Y$ and $g:Y \rightarrow X$ by

$f(x_1, \ldots ,x_{n+1}) =\large( \frac{x_1}{1-x_{n+1}}, \ldots , \frac{x_n}{1-x_{n+1}}),\\g(x_1, \ldots ,x_n)=\large(\frac{2x_1}{{||x||_2}^2+1} ,\ldots ,\frac{2x_n}{{||x||_2}^2+1} , \frac{{||x||_2}^2-1}{{||x||_2}^2+1} )$

Verify that $f$ and $g$ are homeomorphisms such that $f^{-1}=g$.

Show that the graph $G$ of a continuous function $f:X \rightarrow Y$ (with the subspace topology inherited from $X \times Y$ ) is homeomorphic to $X$.

Consider the continuous function $f: \mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}$ by $f(x)=1/x$. The graph of f is the hyperbola $xy=1$. One may conclude from the last exercise that $\mathbb{R} \backslash \{0\}$ is homeomorphic to $xy=1$.

Consider the metric spaces $(X_i,d_i),i=1 ,\cdots ,m$. Check that $d(x,y)=max_id_i(x_i,y_i)$ defined on $X= {\Pi ^m}_{i=1} X_i$ is a metric. The metric topology coincides with the product topology on $X$. This may be concluded from the fact that for $x=(x_1, \cdots ,x_m),\mathbb{B_d}(x,r) ={\Pi ^m}_{i=1} \mathbb{B_{d_i}}(x_i,r)$.

Show that projections $\pi_j:{\Pi^m}_{i=1}X_i \rightarrow X_j$ is continuous.

Verify that $\mathbb{B_b}$ and $\mathbb{B_p}$ are basis for a topology on ${\Pi^m}_\alpha X_\alpha$:

(1)  $\mathbb{B_b}=\{\Pi_{\alpha \in I } U_\alpha :U_\alpha \; is \; open \; in X_\alpha \}$ .

(2) $\mathbb{B_b}=\{\Pi_{\alpha \in I } U_\alpha \in \mathbb{B_b}:U_\alpha=X_\alpha$ for finitely many values of $\alpha\}$.

Suppose that each $X_\alpha$ contains a non-empty proper open subset $U_\alpha$. Show that $\Omega_\mathbb{B_p} = \Omega_\mathbb{B_b}$ iff $I$ has finite cardinality.

Show that the product topology is the smallest topology which makes all projections $\pi_\alpha$ continuous.

Let $f:A \rightarrow \Pi_\alpha X_\alpha$ be given by $f(a)=(f_\alpha(a))$, where the functions $f_\alpha : A \rightarrow X_\alpha$ are given. Suppose $\Pi_\alpha X_\alpha$ has either box topology or product topology. Show that if f is continuous then each $f_\alpha$ is continuous.