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Say we have a linear map, $f: \mathbb{R}^{m+1} \to \mathbb{R}^{n+1}$, and we define $\mathbb{RP}^{n}$ as $(\mathbb{R}^{n+1} - \{0\})/{\sim}$ with $\sim$ define by $x \sim y$ if $y = \lambda x$ for some $\lambda \neq 0 \in \mathbb{R}$.

Then, if we define $[f]$ as $[f]:\mathbb{RP}^{n} \to \mathbb{RP}^{m}$; $[x] \mapsto [f(x)]$ what is a necessary and sufficient condition for this to define a map?

Now, after looking at this, I thought it would just be that $f$ is linear, but that can't be it, since $f$ is stated to be linear. Can anyone clarify this for me?

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Remember that in defining projective space we throw away $0$. –  Qiaochu Yuan Mar 13 '12 at 15:40
So we would need a function f such that f(x) is never 0? –  Mary Mar 13 '12 at 15:52
Well, no, because $f(0) = 0$. But that's okay because we throw away $0$ in the domain as well as the range. –  Qiaochu Yuan Mar 13 '12 at 16:19
I belive that there's a error in the question: @Mary, you mean $[f] \colon \mathbb {RP}^m \to \mathbb{RP}^n$, don't you? –  Giorgio Mossa Mar 13 '12 at 16:40
@you: that's not enough. If $f$ is identically zero, then it never maps to a point that actually lies in projective space. –  Qiaochu Yuan Mar 13 '12 at 17:39

2 Answers 2

To answer your question I'm going to prove something a little more general: I'm going to consider a generic map $f \colon \mathbb R^{m+1} \to \mathbb R^{n+1}$.

The well definition problem for the function is that for every $x_1,x_2 \in X$ such that $[x_1]=[x_2]$ the equality $[f(x_1)]=[f(x_2)]$ holds.

Let $\pi_m \colon \mathbb R^{m+1}\setminus\{0\} \to \mathbb{RP}^m$ be the canonical projection, i.e. the map sending every $x \in \mathbb R^{m+1}\setminus\{0\}$ in $\pi_m(x)=[x] \in \mathbb{RP}^m$. Now the well definition problem can be rephrased in terms of diagrams saying that we want find necessary and sufficient conditions for the commutativity of the diagram below .

enter image description here

This function $[f]$ exists if and only if for each pair $x,y \in \mathbb R^{m+1}\setminus\{0\}$ $[x]=[y]$, i.e such that exists $\lambda \in \mathbb R \setminus \{0\}$ the condition $[f(x)]=\pi_n \circ f(x)=\pi_n \circ f(y)=[f(y)]$, this by universal property of quotient set for a given equivalence relation.

Edit: I'm adding some details about the said universal property for quotient sets.

Let $\pi \colon X \to Y$ be a surjective function. Then for every other function $g \colon X \to Z$ there exists a (necessarily unique) $h \colon Y \to Z$ such that

$$ g = h \circ \pi $$ (i.e. $f$ factors through $\pi$) if and only if for each $x_1,x_2 \in X$ such that $\pi(x_1)=\pi(x_2)$, we have that $g(x_1)=g(x_2)$.

Proof: If this $h$ exists clearly we have that for each pair $x_1,x_2 \in X$ such that $\pi(x_1)=\pi(x_2)$, we have that

$$g(x_1)=h \circ \pi(x_1)=h \circ \pi(x_2) = g(x_2)$$

If $k \colon Y \to Z$ is another function such that $k \circ \pi=g$ then we have the for every $y \in Y$ exists a $x \in X$ such that $\pi(x)=y$, and so

$$h(y)=h \circ \pi(x) = k \circ \pi(x)=k(y)$$

and so $k=h$ (thus if $h$ exists it's unique too).

Let's now show that if the condition holds then function $h$ as above exists. For each $x \in X$ we can consider the set $\left\{g(x') | x' \in X, g(x')=g(x) \right\}$, by the hypothesis this set contains just one element, namely $g(x)$. So its well defined tha map $h(\pi(x))=g(x)$, because $g(x)$ doesn't depend of the choice of the $x \in X$. Clearly this $h$ is the map we were looking for.

Now this theorem can be applied to the problem in the begging letting be $X= \mathbb R^{m+1} \setminus \{0\}$, $Y=\mathbb{RP}^{m}$, $\pi=\pi_m$ , $Z=\mathbb{RP}^{n}$ and $g=\pi_n \circ f$.

In particular if $f \mathbb R^{m+1} \to \mathbb R^{n+1}$ is a linear injective map this we have the map $[f]$ because for every $x_1,x_2 \in \mathbb R^{m+1} \setminus \{0\}$ such that exists $\lambda \in \mathbb R \setminus \{0\}$ for which $x_2=\lambda x_1$ we have that $$f(x_2) = f(\lambda x_1)=\lambda f(x_1)$$ and thus $$\pi_n \circ f(x_2) = \pi_n (\lambda f(x_1)) = \pi_n \circ f(x_1)\ \text{.}$$

Injectivity is required because otherwise we could have points of $\mathbb R^{m+1} \setminus \{0\}$ which were send to $0$, which doesn't belong to $\mathbb R^{n+1} \setminus \{0\}$ so $[f]$ could not exists in this case.

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Hope this explaination isn't to boring –  Giorgio Mossa Mar 14 '12 at 10:18

Such an $f$ existis iff $$f(\mathbb R^m \setminus 0) \subseteq \mathbb R^n \setminus 0 $$ and for $x=\lambda y$ it must follow that $f(x)=\mu f(y)$ for some $\mu$. Therefore $f$ must map lines to sets contained in a line. Any injective linear map will do.

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