My book is Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott of which An Introduction to Manifolds by Loring W. Tu is a prequel.

The characterization of the closed Poincaré dual is given here (the "(5.13)") in Section 5.5. This has $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left.

Question: Why is it $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left?

• See below for why I think $\eta_S$ should be on the left rather than right.

• Previously, in Section 5.3, we had this equivalent definition (the "Lemma") of a nondegenerate pairing between two finite-dimensional vector spaces and Poincaré duality (the "(5.4)").

• Note: I believe the characterization for compact Poincaré dual for compact $S$ and $M$ of finite type is correct with $\eta_S'$ on the right.

• Guess: Could have something to do with sign commutativity of Mayer-Vietoris, as described in Lemma 5.6.

• Guess: Poincare dual as described is indeed with $\eta_S$ on the left, but there's also a unique cohomology class $[\gamma_S]$ that's on the right given by $[\gamma_S] = [-\eta_S]$.

How I got $\int_M \eta_S \wedge \omega$ instead of $\int_M \omega \wedge \eta_S$:

I use $()^{\vee}$, instead of $()^{*}$, to denote dual just like in Section 3.1 of the prequel.

1. Let $\varphi$ be the "linear functional on $H^{k}_cM$" given here.

• Such $\varphi: H^{k}_cM \to \mathbb R$ is given by $\varphi[\omega] = \int_S \iota^{*}\omega$ for $[\omega] \in H^k_cM$ and $\iota: S \to M$ inclusion.

2. Let $\delta$ be the isomorphism of Poincaré duality (the "(5.4)").

• Such $\delta: H^{n-k}M \to (H^{k}_cM)^{\vee}$ is given by $\delta([\tau]) = \delta_{[\tau]}$, for $[\tau] \in H^{n-k}M$ and $\delta_{[\tau]}$ given below.

• $\delta_{[\tau]}([\omega]) = \int_M (\tau \wedge \omega)$, for $[\omega] \in H^k_cM$, under the well-definedness described in Section 24.4 of the prequel (which I think is the full details of the "Because the wedge product is an antiderivation, it descends to cohomology" here) and under the pairing given here, which I believe puts $\tau$ on the left rather than right.

3. $[\eta_S]$ is the inverse image of $\varphi$ under $\delta$.

• By choosing $[\tau] = [\eta_S]$, we get $\delta([\eta_S]) = \delta_{[\eta_S]} = \varphi$, that is, for all $[\omega] \in H^k_cM$,

$$\int_M (\eta_S \wedge \omega) = \int_S \iota^{*}\omega,$$

where $\eta_S$ is on the left rather than right.

My book is Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott of which An Introduction to Manifolds by Loring W. Tu is a prequel.

The characterization of the closed Poincaré dual is given here (the "(5.13)") in Section 5.5. This has $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left.

Question: Why is it $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left?

• See below for why I think $\eta_S$ should be on the left rather than right.

• Previously, in Section 5.3, we had this equivalent definition (the "Lemma") of a nondegenerate pairing between two finite-dimensional vector spaces and Poincaré duality (the "(5.4)").

• Note: I believe the characterization for compact Poincaré dual for compact $S$ and $M$ of finite type is correct with $\eta_S'$ on the right.

• Guess: Could have something to do with sign commutativity of Mayer-Vietoris, as described in Lemma 5.6.

• Guess: Poincare dual as described is indeed with $\eta_S$ on the left, but there's also a unique cohomology class $[\gamma_S]$ that's on the right given by $[\gamma_S] = [-\eta_S]$.

How I got $\int_M \eta_S \wedge \omega$ instead of $\int_M \omega \wedge \eta_S$:

I use $()^{\vee}$, instead of $()^{*}$, to denote dual just like in Section 3.1 of the prequel.

1. Let $\varphi$ be the "linear functional on $H^{k}_cM$" given here.

• Such $\varphi: H^{k}_cM \to \mathbb R$ is given by $\varphi[\omega] = \int_S \iota^{*}\omega$ for $[\omega] \in H^k_cM$ and $\iota: S \to M$ inclusion.

2. Let $\delta$ be the isomorphism of Poincaré duality (the "(5.4)").

• Such $\delta: H^{n-k}M \to (H^{k}_cM)^{\vee}$ is given by $\delta([\tau]) = \delta_{[\tau]}$, for $[\tau] \in H^{n-k}M$ and $\delta_{[\tau]}$ given below.

• $\delta_{[\tau]}([\omega]) = \int_M (\tau \wedge \omega)$, for $[\omega] \in H^k_cM$, under the well-definedness described in Section 24.4 of the prequel (which I think is the full details of the "Because the wedge product is an antiderivation, it descends to cohomology" here) and under the pairing given here, which I believe puts $\tau$ on the left rather than right.

3. $[\eta_S]$ is the inverse image of $\varphi$ under $\delta$.

• By choosing $[\tau] = [\eta_S]$, we get $\delta([\eta_S]) = \delta_{[\eta_S]} = \varphi$, that is, for all $[\omega] \in H^k_cM$,

$$\int_M (\eta_S \wedge \omega) = \int_S \iota^{*}\omega,$$

where $\eta_S$ is on the left rather than right.

Edit: After doing some thinking (it's easier to think when you know something is right/wrong as opposed to thinking about whether or not it's right/wrong, I believe), along with comments of Najib Idrissi and answer of Prof Tu, I think I've got it. Is this right?

We get a unique class $[\gamma_S]$ where for $\gamma_S \in [\gamma_S]$ (or any other element of $[\gamma_S]$), we have that for all $[\omega] \in H^k_cM$ $\omega \in [\omega]$ (or any other element of $[\omega]$),

$$\int_M (\gamma_S \wedge \omega) = \int_M ((-1)^{k} (-1)^{n-k}\omega \wedge \gamma_S) = \int_M (\omega \wedge (-1)^{k} (-1)^{n-k} \gamma_S) = \int_S \iota^{*}\omega$$ and then define $[\eta_S] := (-1)^{k} (-1)^{n-k} [\gamma_S] := [(-1)^{k} (-1)^{n-k} \gamma_S]$.

In this case, I think $[\eta_S] := - [\gamma_S] := [-\gamma_S]$ is a different definition from the one in the preceding paragraph unless $k(n-k)$ is an odd integer or something.