how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$

Recall that we have a map $\Phi : SU(2) \to SO(3)$ that sends a matrix $A \in SU(2)$ to the linear endomorphism $\Phi_A$ on $\Bbb{R}^3$ defined as follows: recall that the Lie algebra $\mathfrak{su}(2)$ has basis given by

$$ E_1 = \left(\begin{array}{cc} 0 & i \\ i & 0 \end{array}\right),\ \ \ E_2 = \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right),\ \ \ E_3 = \left(\begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right).$$

This basis is orthonormal with respect to inner product $\langle U,V\rangle = \frac{1}{2}\textrm{Tr}(UV^*)$ on $\mathfrak{su}(2)$, so we can identify $\mathfrak{su}(2)$ and $\Bbb{R}^3$ as an inner product space. Now, for any $U \in \mathfrak{su}(2)$ we define $\Phi_A(U)=AUA^{-1}$. Then $\Phi_A$ preserves the given inner product on $\mathfrak{su}(2)$, which means $\Phi_A$ is an orthogonal linear mapping of $\mathfrak{su}(2)\simeq \Bbb{R}^3$, by the above identification. Thus, $\mathrm{Im}\ \Phi\subset O(3)$.

Recall that an arbitrary element of $SU(2)$ looks like

$$A = \left(\begin{array}{cc} \alpha & -\overline{\beta} \\ \beta & \overline{\alpha} \end{array}\right)$$

for $\alpha,\beta \in \Bbb{C}$ satisfying $|\alpha|^2 + |\beta|^2 = 1$. Now it is easy to see that $SU(2)$ is homeomorphic to $S^3$, which means in particular that $SU(2)$ is connected. Therefore its image under the continuous group homomorphism $\Phi$ is necessarily contained in the identity component of $O(3)$, which is $SO(3)$.

Again appealing to the coordinate expression for $SU(2)$ given above, one can directly compute the kernel of $\Phi$, that is, find the matrices $A$ in $SU(2)$ such that $AE_iA^{-1} = E_i$ for $i = 1,2,3$. Eventually, one finds $\textrm{Ker}\ \Phi= \{\pm I\}$.

We now show that $\Phi$ is surjective. Suppose you know that $\Phi$ is a covering map. Then there is $U$ open about the identity $I \in SO(3)$ and $V$ open about the identity $I$ in $SU(2)$ such that $U$ is homeomorphic to $V$.

Let $\langle U \rangle$ denote the subgroup generated by the set $U$. Then $\langle U \rangle$ must be contained in $\textrm{Im} \Phi$ that is a subgroup. However because $U$ is open it is also closed using an argument via cosets and connectedness of $SO(3)$ now implies that $SO(3) \subseteq \textrm{Im}\Phi$. It follows that $\Phi$ is surjective and so we have proven that

$$SU(2)/\{ \pm I\} \cong SO(3).$$

$\hspace{6in} \square$

I should say that there is also another argument to prove that $\Phi$ is surjective. Namely, we already know that the associated Lie algebra homomorphism

$$\phi : \mathfrak{su}(2) \longrightarrow \mathfrak{so}(3)$$

is an isomorphism, so it would suffice to prove that the map

$$\textrm{exp} : \mathfrak{so}(3) \longrightarrow SO(3)$$

is surjective.

$SU(2)$ is isomorphic to unit quaternions. In this this article you can find a way to represent every quaternion with a rotation of $\mathbb{R}^3$, namely an element of $SO(3)$, and for every rotation of $\mathbb{R}^3$ there's a quaternion that represents it. So you get a surjective function from quaternions to $SO(3)$, this function should be a homomorphism. Composing you get a surjective homomorphism from $SU(2)$ to $SO(3)$. Check if the kernel is $\{1,-1 \}$.

Maybe this can help you. Check for details :)

The connection between SO(3) and SU(2) is particularly well and simply described, using quaternions as Lorban suggested above, in the following article, https://en.wikipedia.org/wiki/Rotation_group_SO(3)?wteswitched=1#Using_quaternions_of_unit_norm

The 2:1-nature of the map between quaternions and SO(3) results from the use of conjugation, i. e. p ↦ $q p q^{-1}$ , to describe the rotation of p $\in\Bbb{R}^3$, by q $\in\Bbb{H}$ and the fact that both "q" and "-q" define the same rotation of p.