$SU(2)=\{M\in GL(2,\mathbf{C}):det(M)=1,{M}^{\u2020}M=I\}$ is the group of 2by2 complex matrices which are unitary and have determinant 1 (special unitary group). Here ${M}^{\u2020}={\overline{M}}^{T}$ .
Example: SU(2) to SO(3)
The group SU(2)
I want to talk about a particular smooth homomorphism $SU(2)\to SO(3)$ .
We can alternatively characterise $SU(2)$ as the group of matrices $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill \overline{b}\hfill & \hfill \overline{a}\hfill \end{array}\right):a,b\in \mathbf{C},a{}^{2}+b{}^{2}=1\}$ . To see this, note that unitarity implies ${M}^{\u2020}={M}^{1}$ , so $$\left(\begin{array}{cc}\hfill \overline{a}\hfill & \hfill \overline{c}\hfill \\ \hfill \overline{b}\hfill & \hfill \overline{d}\hfill \end{array}\right)=\frac{1}{detM}\left(\begin{array}{cc}\hfill d\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill a\hfill \end{array}\right),$$ and because $det(M)=1$ we get $c=\overline{b}$ and $d=\overline{a}$ . The condition $det(M)=1$ then becomes ${a}^{2}+{b}^{2}=1$ .
The Lie algebra su(2)
The Lie algebra is $$\U0001d530\U0001d532(2)=\{X\in \U0001d524\U0001d529(2,\mathbf{C}):\mathrm{Tr}(X)=0,{X}^{\u2020}=X\}$$ (these conditions come from differentiating $det(\mathrm{exp}(tX))=1$ and $\mathrm{exp}{(tX)}^{\u2020}\mathrm{exp}(tX)=I$ with respect to $t$ at $t=0$ as usual). We can write down a general element of this Lie algebra as follows:$$\U0001d530\U0001d532(2)=\{\left(\begin{array}{cc}\hfill ix\hfill & \hfill y+iz\hfill \\ \hfill y+iz\hfill & \hfill ix\hfill \end{array}\right):x,y,z\in \mathbf{R}\}$$ The fact that the diagonal elements are imaginary follows from ${X}^{\u2020}=X$ . They sum to zero because the trace is zero. The bottomleft entry is determined by the topright by ${X}^{\u2020}=X$ .
Even though $SU(2)$ is a group of complex matrices, $\U0001d530\U0001d532(2)$ is a real vector space, not a complex vector space. That's because $SU(2)$ is cut out by a single real equation (rather than something holomorphic).
If $(v=(x,y,z)\in {\mathbf{R}}^{3}$ , let's write ${M}_{v}=\left(\begin{array}{cc}\hfill ix\hfill & \hfill y+iz\hfill \\ \hfill y+iz\hfill & \hfill ix\hfill \end{array}\right)$ . Here are some useful equations we will use.
$[{M}_{u},{M}_{v}]={M}_{2u\times v}$ and $\mathrm{Tr}({M}_{u}{M}_{v})=2u\cdot v$ .
A homomorphism
If $B\in SU(2)$ then $B{M}_{v}{B}^{1}\in \U0001d530\U0001d532(2)$ for any ${M}_{v}\in \U0001d530\U0001d532(2)$ . This means that $B{M}_{v}{B}^{1}={M}_{w}$ for some $w\in {\mathbf{R}}^{3}$ . Moreover, $w=R(B)v$ for some rotation $R(B)$ . In other words, we have a map $R:SU(2)\to SO(3)$ such that ${M}_{R(B)v}=B{M}_{v}{B}^{1}$ (we will only check that $R(B)$ is an orthogonal matrix, not that it's a rotation).
To see that $B{M}_{v}{B}^{1}\in \U0001d530\U0001d532(2)$ , we need to show that it is tracefree and antiHermitian:

Because trace is conjugation invariant, we have $\mathrm{Tr}(B{M}_{v}{B}^{1})=\mathrm{Tr}({M}_{v})=0$ .

We have ${(B{M}_{v}{B}^{1})}^{\u2020}={({B}^{1})}^{\u2020}{M}_{v}^{\u2020}{B}^{\u2020}=B({M}_{v}){B}^{1}=B{M}_{v}{B}^{1}$ because $B\in SU(2)$ and ${M}_{v}\in \U0001d530\U0001d532(2)$ .
Now define $R(B):{\mathbf{R}}^{3}\to {\mathbf{R}}^{3}$ implicitly by ${M}_{R(B)v}=B{M}_{v}{B}^{1}$ . We want to show that $R(B)$ is an orthogonal matrix, i.e. that it preserves dot products.
Using our formula from before, we have: $$(R(B){v}_{1})\cdot (R(B){v}_{2})=\frac{1}{2}\mathrm{Tr}({M}_{R(B){v}_{1}}{M}_{R(B){v}_{2}}),$$ which is equal to $$\frac{1}{2}\mathrm{Tr}(B{M}_{{v}_{1}}{B}^{1}B{M}_{{v}_{2}}{B}^{1})=\mathrm{Tr}({M}_{{v}_{1}}{M}_{{v}_{2}})$$ because the central ${B}^{1}B$ cancels and the trace is unchanged by conjugation. This is then equal to ${v}_{1}\cdot {v}_{2}$ by the formula from above.
We have therefore defined a map $R:SU(2)\to O(3)$ . This turns out to be a homomorphism. Let's compute ${R}_{*}$ . We know that $R(\mathrm{exp}(t{M}_{u}))=\mathrm{exp}(t{R}_{*}({M}_{u}))$ , so $${R}_{*}({M}_{u})={\frac{d}{dt}}_{t=0}R(\mathrm{exp}(t{M}_{u})).$$
We have ${M}_{R(B)v}=B{M}_{v}{B}^{1}$ , so if $B=\mathrm{exp}(t{M}_{u})$ then we get $${M}_{R(\mathrm{exp}(t{M}_{u})v)}=\mathrm{exp}(t{M}_{u}){M}_{v}\mathrm{exp}(t{M}_{u}).$$ Differentiating with respect to $t$ gives $${M}_{{R}_{*}({M}_{u})v}={\frac{d}{dt}}_{t=0}={M}_{u}{M}_{v}{M}_{v}{M}_{u}$$ by the product rule. Therefore ${M}_{{R}_{*}({M}_{u})v}=[{M}_{u},{M}_{v}]={M}_{2u\times v}$ , so $${R}_{*}({M}_{u})v=2u\times v.$$
Therefore ${R}_{*}({M}_{u})$ is the map which sends $v$ to $2u\times v$ . If we write $u=({u}_{1},{u}_{2},{u}_{3})$ and $v=({v}_{1},{v}_{2},{v}_{3})$ , note that $$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill {u}_{3}\hfill & \hfill {u}_{2}\hfill \\ \hfill {u}_{3}\hfill & \hfill 0\hfill & \hfill {u}_{1}\hfill \\ \hfill {u}_{2}\hfill & \hfill {u}_{1}\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{c}\hfill {v}_{1}\hfill \\ \hfill {v}_{2}\hfill \\ \hfill {v}_{3}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill {u}_{2}{v}_{3}{u}_{3}{v}_{2}\hfill \\ \hfill {u}_{3}{v}_{1}{u}_{1}{v}_{3}\hfill \\ \hfill {u}_{1}{v}_{2}{u}_{2}{v}_{1}\hfill \end{array}\right)=u\times v,$$ so we have $${R}_{*}\left(\begin{array}{cc}\hfill ix\hfill & \hfill y+iz\hfill \\ \hfill y+iz\hfill & \hfill ix\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 2z\hfill & \hfill 2y\hfill \\ \hfill 2z\hfill & \hfill 0\hfill & \hfill 2x\hfill \\ \hfill 2y\hfill & \hfill 2x\hfill & \hfill 0\hfill \end{array}\right).$$
This is a nontrivial example of a Lie algebra homomorphism $\U0001d530\U0001d532(2)\to \U0001d52c(3)$ . Even though it's nontrivial, it is easy to write down: it's linear in $x$ , $y$ , and $z$ .
Concluding remarks
We have this homomorphism $R:SU(2)\to O(3)$ . This is not an isomorphism. For a start, it only hits the rotations, i.e. the orthogonal transformations with determinant $1$ ($SO(3)\subset O(3)$ ). Even if you think of it as a map $SU(2)\to SO(3)$ , it's still not an isomorphism: it is 2to1. For example, $B$ and $B$ both map to the same rotation: $$B{M}_{v}{B}^{1}=(B){M}_{v}{(B)}^{1}.$$
At the level of Lie algebras, ${R}_{*}:\U0001d530\U0001d532(2)\to \U0001d530\U0001d52c(3)$ is an isomorphism: from $\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 2z\hfill & \hfill 2y\hfill \\ \hfill 2z\hfill & \hfill 0\hfill & \hfill 2x\hfill \\ \hfill 2y\hfill & \hfill 2x\hfill & \hfill 0\hfill \end{array}\right)$ we can figure out $x$ , $y$ and $z$ by dividing by $2$ , so we know which matrix $\left(\begin{array}{cc}\hfill ix\hfill & \hfill y+iz\hfill \\ \hfill y+iz\hfill & \hfill ix\hfill \end{array}\right)$ to write down. That is, we have an inverse ${({R}_{*})}^{1}:\U0001d530\U0001d52c(3)\to \U0001d530\U0001d532(2)$ . However, it doesn't correspond to a Lie group homomorphism $SO(3)\to SU(2)$ . This is OK because Lie's theorem about exponentiating homomorphisms only applies when the domain is simplyconnected, and the group $SO(3)$ is not simplyconnected.
Preclass exercises
We constructed a homomorphism $R:SU(2)\to O(3)$ . Can you see why $R$ lands in the subset $SO(3)$ of matrices with determinant 1?
Can you prove that $\U0001d530\U0001d532(2)$ is the set of 2by2 matrices $X$ such that ${X}^{T}=X$ and $\mathrm{Tr}(X)=0$ ?