We know that $\operatorname{Spin}(V \oplus V^*)$ is the double cover of $\operatorname {SO}(V\oplus V^*)$ via the map $$\rho: \operatorname {Spin}(V \oplus V^*)\rightarrow \operatorname {SO}(V\oplus V^*)$$ where $\rho(x)(v)=xvx^{-1}$ for $x\in\operatorname {Spin}(V \oplus V^*)$ and $v\in V\oplus V^*$ now take $\rho_*:\pi_1(\operatorname {Spin}(V \oplus V^*))\rightarrow\pi_1(\operatorname {SO}(V\oplus V^*))$ the homomorphism of fundamental groups. I need to see the image of this map is {$1,(\sigma,\sigma)$} where $\sigma$ is the nontrivial element of $\pi_1(\operatorname {SO}(V\oplus V^*))$. so far I could compute that the group $\operatorname {SO}(V\oplus V^*)$ has two connected components each with fundamental group $\mathbb{Z}_2\times\mathbb{Z_2}$ hence $\operatorname {Spin}(V \oplus V^*)$ has twe connected components each with fundamental group $\mathbb{Z_2}$ but can't figure this out how the image of $\rho_*$ can't be $(0,\sigma)$ or $(\sigma,0)$.
I'll be so thankfull if somebody could give me a hint about this computation.