Structure of SL_2 and root coordinates

For (a), work first over an algebraic closure and write $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $ad-bc=1$. If $a\mathrm{\ne }0$, direct multiplication gives $g=u_{-\alpha }(c\mathrm{/}a)\mathrm{diag}(a,a^{-1})u_{\alpha }(b\mathrm{/}a)$. Put $w(a)=u_{\alpha }(a)u_{-\alpha }(-a^{-1})u_{\alpha }(a)=\left(\begin{array}{cc}0& a\\ -a^{-1}& 0\end{array}\right)$; then $w(a)w(-1)=\mathrm{diag}(a,a^{-1})$. If $a=0$, then $c\mathrm{\ne }0$, and $u_{\alpha }(1)g$ has nonzero top-left entry, reducing to the previous case. Thus $U^{\pm }$ generate the smooth group ${\mathrm{S}\mathrm{L}}_2$: a closed subgroup containing them contains every geometric point, hence the whole reduced group. For perfectness, choose $x$ in the algebraic closure with $x^2\mathrm{\ne }1$. The identity $[\mathrm{diag}(x,x^{-1}),u_{\alpha }(t)]=u_{\alpha }((x^2-1)t)$ puts $U^{+}$ in the derived subgroup; conjugation by $n_{\alpha }$ does the same for $U^{-}$. They generate ${\mathrm{S}\mathrm{L}}_2$, and its quotient ${\mathrm{P}\mathrm{G}\mathrm{L}}_2$ is therefore perfect. These equalities of algebraic groups descend to $k$; no assertion of perfectness of the abstract groups of $k$-points is needed.

For (b), matrix multiplication gives $\mathrm{diag}(x,x^{-1})u_{\alpha }(a)\mathrm{diag}(x^{-1},x)=u_{\alpha }(x^{2}a)$ and $\mathrm{diag}(x,x^{-1})u_{-\alpha }(a)\mathrm{diag}(x^{-1},x)=u_{-\alpha }(x^{-2}a)$. These identities hold over every $k$-algebra and are exactly the stated conjugation formulas for $\alpha (t)=x^2$.

For (c), direct multiplication gives $u_{\alpha }(1)u_{-\alpha }(-1)u_{\alpha }(1)=\left(\genfrac{}{}{0px}{}{01}{-10}\right)=n_{\alpha }$, and $n_{\alpha }^2=-I={\alpha }^{\vee }(-1)$ where ${\alpha }^{\vee }:{\mathbf{G}}_m\to T_2$ is the cocharacter $t\mapsto \mathrm{diag}(t,t^{-1})$ with $\chi \circ {\alpha }^{\vee }=\mathrm{id}$. Since $n_{\alpha }\mathrm{\notin }{T}_2$ but $n_{\alpha }$ normalizes $T_2$ (it conjugates $\mathrm{diag}(x,x^{-1})$ to $\mathrm{diag}(x^{-1},x)$), it represents the nontrivial element $s_{\alpha }$ of $W({\mathrm{S}\mathrm{L}}_2,T_2)$; the induced action on characters is $s_{\alpha }(\chi )=\chi -⟨\chi ,{\alpha }^{\vee }⟩\alpha =\chi -2\chi =-\chi$, the reflection in the root $\alpha =2\chi$.

For (d), fppf locally every class in ${\mathrm{P}\mathrm{G}\mathrm{L}}_2$ has a matrix representative whose determinant can be made $1$ by adjoining a square root and rescaling. Thus ${\mathrm{S}\mathrm{L}}_2\to {\mathrm{P}\mathrm{G}\mathrm{L}}_2$ is surjective as a group scheme, with kernel the scalar matrices of determinant $1$, namely ${\mu }_2$. This is a central isogeny, including characteristic $2$. For the universal-cover assertion, work over an algebraic closure and let $H\to {\mathrm{P}\mathrm{G}\mathrm{L}}_2$ be a central isogeny of smooth connected groups with kernel $N$. By [F4], $H$ is reductive and a maximal torus $S$ contains $N$ and maps onto the one-dimensional diagonal torus. Thus $S\cong {\mathbf{G}}_m$ and $N\cong {\mu }_m$ for some $m$. Lift the nontrivial Weyl normalizer point to $H$; it normalizes $S$ and acts by inversion, because this is its induced action on $S\mathrm{/}N$ and the character lattice map is injective of finite index. Centrality makes that inversion trivial on ${\mu }_m$, so its character group $\mathbb{Z}\mathrm{/}m$ is killed by $2$, forcing $m$ to divide $2$. Any central cover of ${\mathrm{S}\mathrm{L}}_2$, when composed with ${\mathrm{S}\mathrm{L}}_2\to {\mathrm{P}\mathrm{G}\mathrm{L}}_2$, therefore has degree at most two; the latter map already has degree two, so the former has degree one and is an isomorphism. This proves that ${\mathrm{S}\mathrm{L}}_2$ is simply connected and is the universal central cover of ${\mathrm{P}\mathrm{G}\mathrm{L}}_2$, exactly the argument of Milne20.31, including characteristic two. Finally, the pair-automorphism theorem cited in the sources identifies automorphisms of $({\mathrm{S}\mathrm{L}}_2,T_2)$ with conjugations by $(N_{{\mathrm{S}\mathrm{L}}_2}(T_2)\mathrm{/}{\mu }_2)(k)$. Such a conjugation either preserves the two root subgroups or interchanges them. After composing with conjugation by $n_{\alpha }$ if necessary, it preserves $U^{+}$ and is a diagonal conjugation, whose parameter is determined by its action on $U^{+}$. Hence its restrictions to $T_2$ and $U^{+}$ determine it.