The image of a base of absolute roots is a base of relative roots $G$ is a connected, reductive group over a field $k$, $S$ is a maximal $k$-split torus of $G$, and $_k \Phi = \Phi(S,G)$ is the set of roots of $S$ in $G$.  Equivalently, $_k \Phi$ is the restriction to $S$ of the set $\Phi = \Phi(T,G)$ of roots of $T$ in $G$, removing any roots which are trivial on $S$ (e.g. see here mathoverflow.net/questions/255283/two-definitions-of-restricted-roots).
This part in Borel, Linear Algebraic Groups, explains the relationship between certain bases of $\Phi$ and $_k \Phi$.

I've been staring at (3) for awhile.  
It is clear to me that $_k\Delta \subseteq j(\Delta)$: by construction, every element of $_k\Phi^+$ is an nonnegative integral combination of elements of $j(\Delta)$, so $j(\Delta)$ must contain the unique base of $_k \Phi^+$, namely $_k \Delta$.  
But I don't understand why $j(\Delta)$ should be contained in $_k \Delta \cup \{0\}$.
 A: That's really strange since Borel and Tits have a detailed and non-obvious proof of $j(\Delta)\subseteq{}_k\Delta\cup\{0\}$ (Proposition 6.8) in their paper. Here is a simplification which I extracted from the corresponding statement (Prop. 15.5.3) in Springer's book.
There are two $\Gamma=Gal(K/k)$-actions on $X^*(T)$: the natural one $\gamma(\chi)$ and the $*$-action $w*\chi=w_\gamma(\gamma(\chi))$ where $w_\gamma$ is the unique Weyl group element with $\gamma(B)=w_\gamma^{-1}Bw_\gamma$. The key observation is that $w_\gamma$ is an element of the Weyl group of the anisotropic kernel $M=C_G(S)$. Let $\Delta^0\subseteq\Delta$ be the set of simple roots of $M$. Its elements are the simple roots restricting to $0$ on $S$. It follows that
$$
(1)\qquad\gamma(\chi)-\gamma*\chi\in\langle\Delta^0\rangle_{\mathbb Q}.
$$
Let $Y(T):=X_*(T)\otimes\mathbb Q$. Then $Y(S)=Y(T)^\Gamma$ with respect to the natural $\Gamma$-action. Using $(1)$ one sees that $Y(S)$ is defined by the equations
$$
(2)\qquad\{\sigma=0\mid\sigma\in\Delta^0\}\cup\{\chi-\eta=0\mid\chi\in X^*(T),\eta\in\Gamma*\chi\}.
$$
So these span the kernel of $j$. Let $\Delta'\subseteq\Delta\setminus\Delta^0$ be a set of representatives of the $\Gamma*$-orbits. Then $\Delta'$ is linearly independent modulo $(2)$. Thus $j(\Delta)\setminus\{0\}=j(\Delta')$ is linearly independent.
