Taking the point of view that a Thom spectrum functor should be a map $Top_{/BGL_1(R)}\to LMod_R$, for $R$ some $\mathbb{E}_n$-ring spectrum, there seem to be two morphisms (in $Top_{/BGL_1(R)}$) that could potentially be called the "Thom diagonal."

The first (given as diagram 5.43 on page 42 of ABGHR) is given by the commutative triangle (hence the morphism in $Top_{/BGL_1(R)}$):

$$\require{AMScd} \begin{CD} X@>{\Delta}>> X\times X;\\ @V{f}VV @VVf\circ\pi_1V \\ BGL_1(R)@>{=}>> BGL_1(R); \end{CD}$$

and the second (given on page 37 of this other paper by Ando, Blumberg and Gepner) is given in $GL_1(R)$-spaces (equivalent to spaces over $BGL_1(R)$):

$$P\overset{1_P\times \ast}\longrightarrow P\times (GL_1(R)\times X).$$

These seem like different morphisms (i.e. if we use the Grothendieck construction to compare them), are they? My motivation for thinking the latter is not the same as the former is that the latter is equivalent to a morphism that would factor as $$P\overset{1_P\times \ast}\longrightarrow P\times GL_1(R)\overset{1_{P\times GL_1(R)}\times\ast}\longrightarrow P\times (GL_1(R)\times X),$$ but this, on the level of spaces over $BGL_1(R)$, should correspond not to the diagonal map but rather the inclusion into the left coordinate $X\overset{i_1}\hookrightarrow X\times X$ (I think?!).

If this latter map is *NOT* the one coming from the former, then what map in $GL_1(R)$-spaces does come from the former map? If the latter map is the same as the former, how do we see that?