SU(2) equals the set of invertible 2by2 complex matrices M such that det(M)=1 and M dagger M equals the identity is the group of 2by2 complex matrices which are unitary and have determinant 1 (special unitary group). Here M dagger equals M conjugate transpose.
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 a,b,minus b bar, a bar such that a and b are complex numbers and absolute value of a squared + absolute value of b squared equals 1. To see this, note that unitarity implies M dagger equals M inverse, so bar a, bar c, bar b, bar d equals one over det M times d, minus b, minus c, a, and because det M equals 1 we get c equals minus b bar and d equals a bar. The condition det M equals 1 then becomes absolute value of a squared + absolute value of b squared equals 1.
The Lie algebra su(2)
The Lie algebra is little s u 2 equals the set of 2by2 complex matrices X such that trace X equals zero and X dagger equals minus X (these conditions come from differentiating det exp t X equals 1 and (exp t X) dagger times exp t X equals the identity with respect to t at t=0 as usual). We can write down a general element of this Lie algebra as follows:little s u 2 equals the set of matrices i x, y + i z, minus y + i z, minus i x, with x,y,z real numbers. The fact that the diagonal elements are imaginary follows from X dagger equals minus X. They sum to zero because the trace is zero. The bottomleft entry is determined by the topright by X dagger equals minus X.
Even though SU(2) is a group of complex matrices, little s u 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 equals the vector x,y,z in R 3, let's write M_v equals the 2by2 matrix i x, y + i z, minus y + i z, minus i x. Here are some useful equations we will use.
M_u bracket M_v equals M 2 u cross v and trace of M_u times M_v equals minus 2 u dot v.
A homomorphism
If B is in SU(2) then B M_v B inverse is in little s u 2 for any M_v in little s u 2. This means that B M_v B inverse equals M_w for some w in R 3. Moreover, w equals R of B applied to v for some rotation R of B. In other words, we have a map R from SU(2) to SO(3) such that M subscript (R(B) of v) equals B M_v B inverse (we will only check that R of B is an orthogonal matrix, not that it's a rotation).
To see that B M_v B inverse is in little s u 2, we need to show that it is tracefree and antiHermitian:

Because trace is conjugation invariant, we have trace of B M_v B inverse equals the trace of M_v, which equals 0.

We have B M_v B inverse, all dagger, equals B inverse dagger M_v dagger B dagger, which equals minus B times M_v times B inverse because B is in SU(2) and M_v is in little s u 2.
Now define R of B from R 3 to R 3 implicitly by M subscript (R of B v) equals B M_v B inverse. We want to show that R of B is an orthogonal matrix, i.e. that it preserves dot products.
Using our formula from before, we have: R of B v_1 dot R of B v_2 equals minus a half trace of (M subscript (R of B v_1) times M subscript (R of B v_2)) which is equal to minus a half trace of B M v_1 B inverse B M v_2 B inverse, which equals the trace of M v_1 M v_2 because the central B inverse B cancels and the trace is unchanged by conjugation. This is then equal to v_1 dot v_2 by the formula from above.
We have therefore defined a map R from SU(2) to O(3). This turns out to be a homomorphism. Let's compute R star. We know that R of exp t M_u equals exp of t R star M_u, so R star M_u equals d by d t at t=0 of R exp t M_u.
We have M subscript (R of B v) equals B M_v B inverse, so if B equals exp t M_u then we get M subscript (R (exp t M_u) v) equals exp t M_u times M_v times exp of minus t M_u Differentiating with respect to t gives M subscript (R star M_u v) equals d by d t at t=0 of M_u M_v minus M_v M_u by the product rule. Therefore M subscript(R star M_u v) equals M_u bracket M_v, which equals M 2 u cross v, so R star M_u v equals 2 u cross v
Therefore R star M_u is the map which sends v to 2 u cross v. If we write u=(u_1,u_2,u_3) and v=(v_1,v_2,v_3), note that the 3by3 matrix 0, minus u_3, u_2, u_3, 0, minus u_1, minus u_2, u_1, 0 applied to the vector v equals the vector u_2 v_2 minus u_3 v_2, u_3 v_1 minus u_1 v_3, u_1 v_2 minus u_2 v_1, which equals u cross v so we have R star of the matrix ix, y + i z, minus y + i z, minus i x equals the 3by3 matrix 0, minus 2 z, 2 y, 2 z, 0, minus 2 x, minus 2 y, 2 x, 0.
This is a nontrivial example of a Lie algebra homomorphism little s u 2 to little o 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 from 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) inside 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 minus B both map to the same rotation: B M_v B inverse equals minus B M_v minus B inverse.
At the level of Lie algebras, R star little s u 2 to little s o 3 is an isomorphism: from the 3by3 matrix 0, minus 2 z, 2 y, 2 z, 0, minus 2 x, minus 2 y, 2 x, 0. we can figure out x, y and z by dividing by 2, so we know which matrix the matrix ix, y + i z, minus y + i z, minus i x to write down. That is, we have an inverse R star inverse from little s o 3 to little s u 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 from 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 little s u 2 is the set of 2by2 matrices X such that X transpose equals minus X and trace of X equals 0?