The special unitary group is defined as $$\mathrm{SU}(2) = \{A\in M_{2\times 2}(\mathbb{C}) \mid A(\overline{A})^T=I_{2\times2}\}.$$ Show that this is homeomorphic to the $3$-sphere $$\mathbb{S}^3 = \{(a,b)\in \mathbb{C}^2\mid |a|^2+|b|^2=1\}.$$
I know that for a map to be a homeomorphism the function needs to be continuous, function is bijective and the inverse of this function is continuous. However, I am not sure how to show that two things are homeomorphic.
Two topological spaces $X,Y$ are said to be homeomorphic if there is a function $f:\:X\to Y$ which is a homeomorphism.
So all you got to do is find one. Usually there is a natural way to define such function, and that is the case right here : think of a general matrix in $SU2$ - how does it look? how many parameters, or values, define it? where would you send it then?
If needed i'll give a specific solution.
p.s
per your request i'll add more details:
every $M\in SU(2)$ can be desribed as $\begin{bmatrix} a & -\overline b \\ b & \overline a \end{bmatrix}$, and every $a,b$ describes a unique $M$. therefore send $f(M)=(a,b)$. this is clearly a bijection (the inverse is $(a,b)\mapsto \begin{bmatrix} a & -\overline b \\ b & \overline a \end{bmatrix}$ , and I'll leave it to you to try to show that is a homeomorphism. Again, if required, i'll assist.
p.s.s
Per your request, a prove that every matrix is of that form:
Let $M=\begin{bmatrix} a & c \\ b & d \end{bmatrix}\in U$. we have $M\overline M^t$= \begin{bmatrix} aa^*+cc^* & ab^*+cd^* \\ ba^*+dc^* & bb^*+dd^* \end{bmatrix}, and we must have $M\overline M^t=I $. the solution for these 4 equations is $c=-\overline b$ and $d=\overline a$, please check yourself.