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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 $0$.2012-03-13
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    So we would need a function f such that f(x) is never 0?2012-03-13
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    Well, no, because $f(0) = 0$. But that's okay because we throw away $0$ in the domain as well as the range.2012-03-13
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    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?2012-03-13
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    It needs to be _well-defined_, meaning that if we choose two representatives of the same class, $[f]$ had better send them to the same place, i.e. if $[x]=[y]$ we have to have $[f(x)]=[f(y)]$. In other words $\forall x\in \mathbb{R}^{m+1}\setminus0,\ \forall \lambda\in\mathbb{R}\setminus0\,\ f(\lambda x)=\lambda' f(x)$ for some $\lambda'$. But since $f$ is linear by assumption, $f(\lambda x)=\lambda f(x)$. So I understand your consternation, it seems that $f$ being linear already implies the induced map is well-defined.2012-03-13
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    @you: that's not enough. If $f$ is identically zero, then it never maps to a point that actually lies in projective space.2012-03-13

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