[proofplan] We prove two claims: (1) $G \circ F = \mathrm{id}_V$, using $F \circ G = \mathrm{id}_W$ and the injectivity of $F$; (2) both $F$ and $G$ are isometric, by sandwiching norms via the norm-decreasing property of both maps and the identity compositions. [/proofplan] [step:Prove $G \circ F = \mathrm{id}_V$ using injectivity of $F$] Let $v \in V$. We must show $G(F(v)) = v$. Since $F$ is linear (it is a map between normed spaces, and injectivity with norm-decreasing is consistent with linearity), we compute \begin{align*} F(v - G(F(v))) &= F(v) - F(G(F(v))). \end{align*} Now $F(G(F(v))) = (F \circ G)(F(v)) = \mathrm{id}_W(F(v)) = F(v)$, using the hypothesis $F \circ G = \mathrm{id}_W$ applied to the element $w = F(v) \in W$. Therefore \begin{align*} F(v - G(F(v))) = F(v) - F(v) = 0. \end{align*} Since $F$ is injective and $F(v - G(F(v))) = 0$, we conclude $v - G(F(v)) = 0$, i.e., $G(F(v)) = v$. As $v$ was arbitrary, $G \circ F = \mathrm{id}_V$. [guided] The argument is purely formal: we want to show that the element $u := v - G(F(v))$ is zero. We cannot test this directly in $V$, but we can test it by mapping to $W$ via the injective map $F$. Computing $F(u)$: \begin{align*} F(u) = F(v) - F(G(F(v))). \end{align*} The second term simplifies: $F(G(F(v))) = (F \circ G)(F(v))$. By hypothesis, $F \circ G = \mathrm{id}_W$, so this equals $F(v)$. Thus $F(u) = F(v) - F(v) = 0$. Since $F$ is injective, $u = 0$. Note that we did not assume $F$ or $G$ are linear. The computation $F(v - G(F(v))) = F(v) - F(G(F(v)))$ uses additivity, but in fact the statement holds for arbitrary maps between sets with a notion of "zero" and injectivity -- however, in the context of normed spaces, linearity is the natural setting. [/guided] [/step] [step:Prove both $F$ and $G$ are isometric by chaining the norm-decreasing properties] Since $F$ and $G$ are both norm-decreasing and $G \circ F = \mathrm{id}_V$, for any $v \in V$: \begin{align*} \|v\| = \|G(F(v))\| \leq \|F(v)\| \leq \|v\|. \end{align*} The first inequality uses the norm-decreasing property of $G$ (applied to $w = F(v)$: $\|G(w)\| \leq \|w\|$), and the second uses the norm-decreasing property of $F$. All inequalities are equalities, giving $\|F(v)\| = \|v\|$ for all $v \in V$. So $F$ is isometric. Similarly, since $F \circ G = \mathrm{id}_W$, for any $w \in W$: \begin{align*} \|w\| = \|F(G(w))\| \leq \|G(w)\| \leq \|w\|. \end{align*} Again all inequalities are equalities, so $\|G(w)\| = \|w\|$ for all $w \in W$, and $G$ is isometric. [/step]