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.

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.

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**.

$[\eta_S]$ 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.