Why is $q(f,g) = (f-g,0)$ not adjointable?

Let $$A= C([0,1])$$ and $$J= \{f \in A: f(0) = 0\}$$. Consider the Hilbert $$C^*$$-module $$E:= A \oplus J$$ (with the obvious right $$A$$-action and inner product). I want to prove that $$q: E \to E: (f,g) \mapsto (f-g, 0)$$ is not adjointable. This is claimed in Lance's book on Hilbert $$C^*$$-modules, p22.

Here is what I tried. Assume to the contrary that $$q$$ is adjointable. Then there is $$q^*: E \to E$$ such that $$(\overline{f-g})s= \langle q(f,g) , (s,t)\rangle = \langle (f,g), q^*(s,t)\rangle.$$

In particular, $$q^*(s,t)$$ does not depend on $$t$$ so we have $$q^*(s,t) = q^*(s,0)$$. Then I'm stuck.

This is just a calculation. Continuning your argument, $$q^*(s,t) = q^*(s,0) = (s_1,-s_2)$$ for some $$s_1\in A, s_2\in J$$ (I add the minus sign for convenience later). Then $$\overline{f} s - \overline{g}s = \langle (f,g), (s_1,-s_2) \rangle = \overline{f} s_1 - \overline{g}s_2,$$ for all $$f\in A, g\in J$$. Set $$f=1,g=0$$ to see that $$s = s_1$$; set $$f=0$$ to see that $$\overline{g} s = \overline{g} s_2,$$ for all $$g\in J$$. Letting $$g$$ run through an approximate identity for $$J$$ (so a net $$(g_i)$$ with $$g_i(x)\rightarrow 1$$ for each $$x>0$$) we conclude that $$s(x) = s_2(x)$$ for all $$x>0$$. If for example $$s=1\in A\setminus J$$ this shows that $$s_2(x)=1$$ for all $$x>0$$, contradicting that $$s_2\in J$$.