Q:

Maps between real projective spaces

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? 

A:

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 .  

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.