Homeomorphisms of quotient spaces 
Let: $$s_0 \in S^1 - \text{fixed point on the circle,}$$ $$S^2=\left\{(x,y,z)\in \mathbb R^3: x^2+y^2+z^2=1\right\} \text{ - sphere}$$
A) Prove that quotient space $S^1\times I/S^1 \times \left\{1\right\}$ is homeomorphic with hemisphere $S^2_+=\left\{(x,y,z)\in S^2:z \ge 0\right\}$
B) Let $X=\left\{(a,b)\in S^1 \times S^1: a=s_0 \vee  b=s_0\right\}$. Prove that quotient space $S^1\times S^1/X$ is homeomorphic with the sphere $S^2$

In both examples, I tried to find a formula for homeomorphism. But I was unable to find transformations that would meet the properties of homeomorphism. Moreover, I wonder if there is an easier way to prove it than to find specific transformations.Can someone help me and show me how to approach such tasks?
 A: A) Define $f : S^1 \times I \to S^2_+, f(x,y,t) = ((1-t)x,(1-t)y,\sqrt{2t-t^2)})$. We have $f(x,y,t) = f(x',y',t')$ iff $(x,y,t) = (x',y',t')$ or $t= t' =1$. Thus $f$ induces a continuous bijection $F : S^1 \times I / S^1 \times \{1\} \to S^2_+$. But $S^1 \times I / S^1 \times \{1\}$ is compact and $S^2_+$ is Hausdorff, thus $F$ is a homeomorphism.
B) This is more complicated. Let us first observe that $S^1$ is homeomorphic to the quotient space obtained from $[-1,1]$ by identifying $-1$ and $1$. Let $p : [-1,1] \to S^1, p(t) = (\cos \pi t, \sin\pi t)$, and $q : S^1 \times S^1 \to S^1 \times S^1/X$, denote the quotient maps. Consider the map
$$g : [-1,1] \times [-1,1] \stackrel{p\times p}{\rightarrow} S^1 \times S^1 \stackrel{q}{\rightarrow} S^1 \times S^1/X.$$
As a composition of quotient maps it is a quotient map. It is easy to vertify that $g(s,t) = g(s',t')$ iff $(s,t) = (s',t')$ of $(s,t), (s',t') \in \partial ( [-1,1] \times [-1,1]) =  [-1,1] \times \{-1,1\} \cup \{-1,1\} \times [-1,1]$. Thus $g$ induces a continuous bijection $G :  [-1,1] \times [-1,1]/\partial ( [-1,1] \times [-1,1]) \to S^1 \times S^1/X$. This map is again a homeomorphism.
But the pair $([-1,1] \times [-1,1],\partial ( [-1,1] \times [-1,1]))$ is homeomorphic to $(D^2,S^1)$ (where $D^2$ is the closed unit disk in the plane). See my answer to $(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic and take $m=2$, $\lVert - \rVert_1$ = maximum-norm, $\lVert - \rVert_2$ = Euclidean norm.
Thus it remains to show that $D^2/S^1$ is homeomorphic to $S^2$. Let $r = \lVert (x,y) \rVert_2$ and define $$h : D^2  \to S^2, h(x,y) = \begin{cases} \left(2x,2y,\sqrt{1-4r^2} \right) & 0 \le r \le 1/2 \\ \left(4(1 - r)x,4(1 - r)y,-\sqrt{1- 16(1-r)^2r^2} \right) &  1/2 \le r \le 1 \end{cases} .$$
This induces the desired homeomorphism.
