ScalingStacks

Remark 7.3.12. Fix an orientation of each component of $Z$ (forgetting about the already given orientation of $Z_{o})$ and define $Z^{+}\subset T(Z)$ to be the set of pairs $(z,c)$ such that there is an oriented path $\gamma$ in $Z$ (for the given new orientation) with $m_{c}^{+}(\gamma)=1$.

There is a quotient map $L(Z)\to{\mathbf{Z}}^{\pi_{0}(Z)}$ given by $e_{c}\mapsto e_{\Omega}$ for all $s\in\Omega$ and $(z,c)\in Z^{+}$. Let us show that the bilinear form $R(Z)\times R(Z)\to{\mathbf{Z}}^{\pi_{0}(Z)}$ obtained by composing $\langle-,-\rangle$ with this quotient map is antisymmetric. Let $\gamma$ and $\gamma^{\prime}$ be two injective oriented paths in $Z$ (for the given new orientation). If the supports of $\gamma$ and $\gamma^{\prime}$ are disjoint, then $\langle\llbracket\gamma\rrbracket,\llbracket\gamma^{\prime}\rrbracket\rangle=0$. We have $\langle\llbracket\gamma\rrbracket,\llbracket\gamma\rrbracket\rangle=-e_{\gamma(0+)}-e_{\gamma(1-)}$. If $\gamma([0,1])\cap\gamma^{\prime}([0,1])=\{\gamma(1)\}$, then

$$\langle\llbracket\gamma\rrbracket,\llbracket\gamma^{\prime}\rrbracket\rangle=-e_{\gamma^{\prime}(0+)}\text{ and }\langle\llbracket\gamma\rrbracket,\llbracket\gamma^{\prime}\rrbracket\rangle=-e_{\gamma(1-)}.$$

We deduce the antisymmetry statement.