Path-connectedness of SL(2,C)

Write the columns of $A$ as $u,v\in {\mathbb{C}}^2$. Since $\det A=1$, $u$ and $v$ are linearly independent. Put $r_{11}:=\mathrm{\parallel }u\mathrm{\parallel }>0$, $q_1:=u\mathrm{/}{r}_{11}$, $\alpha :=q_1^{†}v$, $w:=v-\alpha q_1$, and $r_{22}:=\mathrm{\parallel }w\mathrm{\parallel }>0$. Then $q_1^{†}w=0$ and $q_2:=w\mathrm{/}{r}_{22}$ makes $Q:=(q_1,q_2)$ unitary; with $R:=\left(\begin{array}{cc}{r}_{11}& \alpha \\ 0& r_{22}\end{array}\right)$ we have $A=QR$. By [F1]–[F2], $1=\mathrm{\mid }\det A\mathrm{\mid }=\mathrm{\mid }\det Q\mathrm{\mid }{r}_{11}{r}_{22}=r_{11}{r}_{22}$, and $\det A=\det Q{r}_{11}{r}_{22}=1$ then gives $\det Q=1$. Therefore $Q\in SU(2)$ and $R=\left(\begin{array}{cc}r& z\\ 0& r^{-1}\end{array}\right)$ for $r>0$ and $z\in \mathbb{C}$.

If the first column of $Q$ is $(a,c{)}^{\mathsf{T}}$, its unit orthogonal vectors are $z(-\bar{c},\bar{a}{)}^{\mathsf{T}}$ with $\mathrm{\mid }z\mathrm{\mid }=1$. The determinant of the matrix with these columns is $z(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2)=z$; $\det Q=1$ therefore forces $z=1$. Thus $Q=\left(\begin{array}{cc}a& -\bar{c}\\ c& \bar{a}\end{array}\right)$ and $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2=1$. This identifies $SU(2)$ with the unit sphere $S^3\subset {\mathbb{R}}^4$. Under this identification use the standard Euclidean inner product $\cdot$ and norm $\parallel \cdot {\parallel }_2$ on ${\mathbb{R}}^4$. The sphere is path-connected: non-antipodal points are joined by normalizing their straight segment. For antipodal points $x,-x\in S^3$, at least one of the first two standard basis vectors $e_1,e_2$ is not parallel to $x$; choose such an $e_j$ and set $w:=\frac{e_j-(e_j\cdot x)x}{\parallel e_j-(e_j\cdot x)x{\parallel }_2}\mathrm{.}$ The denominator is nonzero because $e_j$ is not parallel to $x$; hence $w\in S^3$ and $w\cdot x=0$. The normalized straight segments from $x$ to $w$ and from $w$ to $-x$ give a path. Hence there is a continuous path $Q_t$ in $SU(2)$ from $I$ to $Q$.

For $R=\left(\begin{array}{cc}r& z\\ 0& r^{-1}\end{array}\right)$, the path $R_t:=\left(\begin{array}{cc}{r}^t& tz\\ 0& r^{-t}\end{array}\right)$ has continuous entries because $r^{\pm t}=\exp (\pm t\log r)$ for the fixed $r>0$, and the real exponential is continuous. It has determinant one for every $t\in [0,1]$, with $R_0=I$ and $R_1=R$. Thus $A_t:=Q_t{R}_t$ is a continuous path in $SL(2,\mathbb{C})$ from $I$ to $A=QR$. Since $A$ was arbitrary, $SL(2,\mathbb{C})$ is path-connected.