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In the O'neill's differential geometry text, there are following problems. enter image description here

The projective plane $P$ is defined as follows. enter image description here

As you know, this $P$ is an abstract surface. So, we have to define the tangent vectors on an abstract surface.

But, this text book just gives following definitions.

Def) Let $\alpha:I \rightarrow M$ be a curve on the abstract surface $M$. The velocity vector( or tangent vector) $\alpha'(t)$ is defined by $\alpha'(t)=(f\alpha)'(t)$ for each differentiable function $f:M\rightarrow R$.

I think the answer of (a) is {$v_p, -v_{-p}$} where $v_p$ is a tangent vector on $\Sigma$.

How can we evaluate this problem?

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Let $x$ in $P$, $x=F(y)=F(-y), y\in \Sigma$. Let $v\in T_xP$, since $dF_y$ and $dF_{-y}$ are isomorphism since $F$ is a local diffeomorphism, there exist a unique $v_y\in T_y\Sigma$ and a unique $v_{-y}\in T_{-y}$ such that $dF_y(v_y)=dF_{-y}(v_{-y})=v$.

Since $\Sigma$ is compact and connected and $F$ continue, $F(\Sigma)=P$ is compact and connected.

Suppose that $P$ is orientable and $\omega$ a volume form on $P$, $F^*\omega$ is invariant by $i:x\rightarrow -x$, this implies that $i^*(F^*\omega)=F^*\omega=-F^*\omega$. This implies that $F^*\omega=0$ contradiction.

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  • $\begingroup$ Could you explain why $F$ is a local diffeomorphism? $F$ is defined by $F(p)=${$p,-p$}. $\endgroup$ – Chris kim Jun 1 '16 at 0:59

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