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Definition (projectivity). Let $V,W$ vector spaces over some field $k$ and $V\neq\{0\}$. Let $F\::V\to W$ an injective linear map and let $v \in V\ \setminus\{0\}$. A projective map $\mathbb{P}(F\ ):\mathbb{P}(V\ )\to\mathbb{P}(W\ ):kv\mapsto kF(v)$ is a projectivity if it is bijective.

Problem. Let $s,t \in \mathbb{P}^{1}_\mathbb{R}, s\neq t$. Define $f:\mathbb{P}^{1}_\mathbb{R} \to \mathbb{P}^{1}_\mathbb{R}$ where $f(s)=t, f(t)=s$ and $\forall p \in \mathbb{P}^{1}_\mathbb{R} \setminus \{s,t\} :f(p) = p$. Show that $f$ is not a projectivity.

I've got this as an exercise in my geometry textbook but I'm having trouble proving it. What I am trying to do is assume that $f$ is a projectivity, and then try to obtain the contradiction. I'll show my progress below. A hint would be welcome! If I am not on the right track at all, please do point this out. Thanks. PS: I'll add some things I think are needed in the proof:

  • $kv_i = kv_j \Rightarrow v_i = \lambda v_j$,
  • $s,t, s\neq t \in \mathbb{P}(V) \Rightarrow s,t$ independent,
  • Let $s=kv_s, t=kv_t$ and $s,t$ independent, then $v_s, v_t$ linear independent,
  • Since $F$ injective, $v, w$ linear independent $\Rightarrow F(v), F(w)$ independent.

Progress. It suffices to show that $f$ is not a projective map. Let $s=kv_s, t=kv_t$. Note that $s,t$ independent $\Rightarrow v_s, v_t$ linear independent. Now assume that $f$ is a projective map. Then $f(kv_s) = kF(v_s) =kv_t$ and $f(kv_t) = kF(v_t) = kv_s$. These imply $F(v_s) = \lambda v_t, F(v_t) = \lambda v_s$.

Now I am stuck... I feel I should come to a contradiction where $s,t$ are not independent. Perhaps working with the fact that $f$ is bijective could help here?

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  • $\begingroup$ In you definition, you need $F$ to be an isomorphisim not only injection, i.e. the dimension of $V,W $ should be the same. Right? $\endgroup$ – user65304 Apr 17 '16 at 19:13
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Show that if a projectivity leaves three points fixed, then it is the identity.

Then pick three points which are different from $r$ and $s$ and show they are fixed by your $f$.

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  • $\begingroup$ This surely is taking the proof another way. Could you perhaps edit your comment to contain the proof of your implication? $\endgroup$ – user12205 Dec 10 '11 at 1:09
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    $\begingroup$ Try to prove it and tell us where you get stuck, better! ;) $\endgroup$ – Mariano Suárez-Álvarez Dec 10 '11 at 2:46

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