$\mu_b$ as in equation . By definition of $T_\theta$ on 2-morphisms it suffices to show

$$\begin{aligned}
\langle \psi^*\theta,\nu\rangle = \langle {\psi'}^*\theta,\nu\rangle . \end{aligned}$$

Indeed, if we see $h$ as a map defined on $M\times [0,1]$ we find a chain homotopy between the chain maps $\psi_*$ and $\psi'_*$ given by

$$\begin{aligned}
{\psi_*}_{p}- {\psi'_*}_{p} = \partial H_p + H_{p-1} \partial \quad \text{for all $p\in\mathbb{Z}$}, \end{aligned}$$

where $H_p := {h_*}_{p+1} D_p$ and

$$\begin{aligned}
D_p : S_p(M) {\longrightarrow}S_{p+1}(M\times [0,1]), \quad c {\longmapsto}c\times [0,1] \end{aligned}$$

is defined using the cross-product, see [@bredon IV.16]. Hence,

$$\begin{aligned}
\langle \psi^*\theta,\nu\rangle - \langle {\psi'}^*\theta,\nu\rangle = \langle  \theta, H_{n-1} \partial \nu  \rangle = \langle h^* \theta, \partial\nu \times [0,1] \rangle .\label{eqnhtpinvvanish}\end{aligned}$$

The homotopy $h$ being stationary on the boundary entails $$\begin{aligned}
h|_{\partial M \times [0,1]}= \psi \circ p_{\partial M}\end{aligned}$$ with the projection $p_{\partial M} \colon \partial M \times [0,1] {\longrightarrow}\partial M$. This yields $$\begin{aligned}
\langle h^* \theta, \partial\nu \times [0,1] \rangle = \langle \psi^* \theta, {p_{\partial M}}_* (\partial\nu \times [0,1]) \rangle \end{aligned}$$ We have

$$\begin{aligned}
 \partial \left( {p_{\partial M}}_* (\partial\nu \times [0,1])\right) = 0,\end{aligned}$$

where we use that the boundaries corresponding to the $[0,1]$ part cancel under the projection, i.e. $$\begin{aligned}
{p_X}_* ( X \times \partial [0,1])= 0,\end{aligned}$$ for any 