Unnecessary condition in the definition of an isomorphism of root systems

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Let $R$ be a root system in a vector space $V$ (of characteristic $0$) with coroots $R^{\vee} \subseteq V^*$. Recall that for every $\alpha \in R$ the corresponding coroot $\alpha^{\vee} \in R^{\vee}$ is uniquely determined by $R$ via the conditions $\alpha^{\vee}(\alpha) = 2$ and $\beta - \alpha^{\vee}(\beta) \alpha \in R$ for every $\beta \in R$.

From what I’ve seen an isomorphism of root systems $f \colon (V, R) \to (V', R')$ is usually defined as a vector space isomorphism $V \to V'$ such that $f(R) = R'$ and $$ f(\alpha)^{\vee}( f(\beta) ) = \alpha^{\vee}(\beta) \quad \text{for all $\alpha, \beta \in R$} \,. \tag*{(1)} $$ (This definition can for example be found in [Definition 18.13.1, 1] and in [p. 43, 2].)

Question: Why is condition $(1)$ explicitely required?

It seems to me that this condition automatically holds:

If $f \colon V \to V'$ is any isomorphism of vector spaces (not necessarily with $f(R) = R'$) then we can push forward the root system $R$ of $V$ to a root system $\tilde{R}'$ of $V'$ given by the set $\tilde{R}' = f(R)$ and coroots $f(\alpha)^{\vee} = \alpha^\vee \circ f^{-1}$.

If now $f(R) = R'$ then the root systems $R'$ and $\tilde{R}'$ have the same underlying set, so for $f(\alpha) \in f(R)$ we can consider its coroot with respect to both $R'$ and $\tilde{R}'$. But it follows from the above uniqueness of coroots that these coincide. Thus $f(\alpha)^{\vee}(f(\beta)) = \alpha^{\vee}(f^{-1}(f(\beta))) = \alpha^{\vee}(\beta)$ as required by $(1)$.


[1] Lie Algebras and Algebraic Groups by Tauvel, Yu
[2] Introduction to Lie Algebras and Representation Theory by James E. Humphreys