I've been playing around with orientations of Thom spectra lately and I realized that I kept getting tripped up by the difference between the notion of an $E$-orientation given in [ABGHR] and the "classical" definition given, for instance, in Lewis' work in [LMS, Ch. IX].

The main difference is that Lewis considers Thom spectra which comes from maps $f\colon X\to Pic(\mathbb{S})$ and then asks for them to be $E$-oriented with respect to some $\mathbb{S}$-algebra $E$; while on the other hand, Ando, Blumberg, Gepner, Hopkins and Rezk say that if you've got a map $f\colon X\to Pic(R)$ for some $\mathbb{S}$-algebra $R$ then $Mf$ is $R$-oriented when $f$ is null, i.e. it factors through the one-point space. Ando, et al. then show that this notion is equivalent to asking for a map $p_!f\simeq Mf\to R$ to have an adjoint, $f\to p^\ast(R)$, which is an equivalence of functors. This last part is a bit self-evident since the constant functor is literally the functor $X\to \ast\to RMod$ that picks out $R$. But the equivalence of the first two notions was less clear to me, so I tried to work through it below.

Before I say anything I should say that a bunch of the stuff below can be unwound from the definitions and results of [ACB], but (1) I didn't know that when I started trying to work this out and (2) I think what's below is maybe a bit more explicit and direct. This is also my first time trying to write some math out on functors.net, which I think is a pretty cool framework for writing math blogs (despite having a few bugs).

**Definition 1:**
Let $f\colon X\to Pic(R)$ be a morphism of $\infty$-groupoids for $R$ some commutative ring spectrum. Let $E$ be an $\mathbb{E}_k$-$R$-algebra. Then we say that an $L$-type $E$-orientation of $Mf\simeq colim(X\to Pic(R)\to RMod)$ (the Thom spectrum as defined in [ABGHR]) is the data of an $R$-module map $u\colon Mf\to E$ satisfying the following property for all points $x\in X$:

- Let $Mx$ be the Thom spectrum associated to the inclusion $x\hookrightarrow X\to Pic(R)$ and let $u_x\colon Mx\to E$ by the composition of the induced map $Mx\to Mf$ with $u\colon Mf\to E$. Then the image of $u_x$ under the equivalence $RMod(Mx,E)\simeq EMod(E\otimes Mx,E)$, i.e. the composition $E\simeq Mx\otimes E\to E\otimes E\to E$, which is $u_x\otimes E$ followed by the multiplication of $E$, is an equivalence of $E$-modules

**Definition 2:**
Let $E$ and $R$ be as in Definition 1. Let $Mg$ be the Thom spectrum associated to a map of $\infty$-groupoids $g\colon X\to Pic(E)$. Then an $A$-type $E$-orientation of $Mg$ is a morphism of $E$-modules $u\colon Mg\to E$ with the following property:

- Let $p\colon X\to \ast$ be the terminal map, which induces an adjunction $p_!\colon EMod^X\rightleftarrows EMod\colon p^\ast$, and note that by definition $Mg\simeq p_!(g)$. Then the image of $u$ under the equivalence $EMod(p_!(f),E)\simeq EMod^X(f,p^\ast(E))$ is an equivalence.

**Lemma 3:**
Let $\mathscr{C}$ be be an $\infty$-category with $E\in \mathscr{C}$ and $X$ a simplicial set. Then there is an equivalence of $\infty$-categories $(\mathscr{C}_{/E})^X\simeq (\mathscr{C}^X)_{/E_X}$ where $E_X$ is the constant functor $X\to \mathscr{C}$ taking every point of $X$ to $E$.

**Proof:**
From[RV, 4.2.1] the slice $\infty$-category $\mathscr{C}_{/E}$ is the pullback of the cospan $E\colon 1\to \mathscr{C}\leftarrow \mathscr{C}\colon id_\mathscr{C}$. Note that the identity functor $id_\mathscr{C}\colon\mathscr{C}\to\mathscr{C}$ is an isofibration (in the language of [RV]) and therefore pulling back along it is a cosmological limit by [RV, 1.2.1(i)] (and the ensuing discussion). Therefore by [RV, 1.3.4(iii)] we have that the functor $(-)^X$ preserves that pullback (because $(-)^X$ is a cosmological functor which by [RV, 1.3.1] preserves cosmological limits). Thus $(\mathscr{C}_{/E})^X$ is the pullback of the cospan $1\overset{E_X}\to \mathscr{C}^X\overset{id_{\mathscr{C}^X}}\leftarrow \mathscr{C}^X$ which is the definition of the slice category $(\mathscr{C}^X)_{/E_X}$.

$\hspace{10in}\blacksquare$

**Remark 4:** Since I've implicitly been working in Lurie's framework, and the above proof switches to the Riehl-Verity framework, I should mention that [RV, D.2.16] proves that Lurie's slice construction and Riehl and Verity's slice construction agree.

**Remark 5:**
Recall that, given a natural transformation $\eta\colon F\Rightarrow G$ of functors $F,G\in \mathscr{C}^X$, $\eta$ is an equivalence of functors if and only if the restrictions $\eta_x\colon F_x\Rightarrow G_x$ are equivalences for all $x\in X$. As a result the equivalence of Lemma 3 identifies functors that factor through the full subcategory of $\mathscr{C}_{/E}$ spanned by equivalences to $E$ with natural transformations to $E_X$ which are equivalences.

**Proposition 6:**
Let $f$, $R$ and $E$ be as in Definition 1. Then a map of $R$-modules $\theta\colon Mf\to E$ is an $L$-type $E$orientation if and only its adjoint $\tilde{\theta}\colon Mf\otimes E\to E$ is an $A$-type $E$-orientation.

**Proof:**
First suppose that $\theta\colon Mf\to E$ is an $L$-type orientation for a Thom spectrum associated to a map $f\colon X\to Pic(R)$. Then by the adjunction $RMod(p_!(f),E)\simeq RMod^X(f, p^\ast(E))\simeq RMod^X(f, E_X)$ we have that $\theta$ determines an object of $(RMod^X)_{/E_X}$ (which in a slight abuse of notation we will also call $\theta$). Therefore by Lemma 3 we have a functor $f_{\theta}\colon X\to RMod_{/E}$. By adjunction this determines a functor $\tilde{f}_{\theta}\colon X\to EMod_{/E}$ (or equivalently by applying the free $E$-module functor and then composing with the multiplication of $E$). Note that because colimits in overcategories are computed in the underlying category, the colimit of $\tilde{f}_{\theta}$ is $\tilde{\theta}\colon Mf\otimes E\simeq M(\tilde{f}_\theta)\to E$, the adjoint of the $L$-type orientation $\theta\colon Mf\to E$. Similarly, for each $x\in X$ the induced functor $x\hookrightarrow X\to EMod_{/E}$ is $\tilde{\theta}_x\colon Mx\otimes E\to E$, the adjoint of the restriction of the $L$-type orientation $\theta_x\colon Mx\to E$. By assumption, each morphism $\tilde{\theta}_x$ is an equivalence. Therefore by Lemma 3 and Remark 5 this induces an equivalence of functors $\tilde{f}_\theta\simeq E_X\in EMod^X$. Therefore $\tilde{\theta}$ is an $A$-type orientation of $Mf\otimes E$. The reverse implication is basically the above argument in reverse.

$\hspace{10in}\blacksquare$

**Corollary 7:**
Let $f\colon X\to Pic(R)$ be a morphism of $\infty$-groupoids with associated Thom spectrum $Mf$. Then a map of $R$-modules $\theta\colon Mf\to E$ is an $L$-type $E$-orientation if and only if its adjoint $\tilde{\theta}\colon Mf\otimes E\to E$ is an $A$-type $E$-orientation if and only if the composite $X\overset{f}\to Pic(R)\overset{-\otimes E}\to Pic(E)$ is null.

**Proof:**
This follows from [ABGHR, 2.24] which identifies $A$-type $E$-orientations with nullhomotopies.

$\hspace{10in}\blacksquare$

**References:**

[ABGHR] = https://arxiv.org/pdf/1403.4325.pdf

[ACB] = https://arxiv.org/pdf/1411.7988.pdf