On when quotients by a group action is a fibration or not, grassmanians Let $V$ be a vector space over $\mathbb R$ with $\text{dim }V\geq k$ and let $M = \text{Mon}(\mathbb R^k, V)$ denote the space of monomorphisms (injective linear maps) of $\mathbb R^k$ into $V$. 
Then there is a quotient map $M \rightarrow Gr(k,V)$. Is this a fibration? I think the map $M \rightarrow Gr(k,V)$ is equivalent to the quotient of $\text{Mon}(\mathbb R^k, V)$ by $GL(k)$ acting on the right by precomposition? I can't find any info on when quotients of smooth manifolds by a smooth action of a Lie group is a fiber bundle/fibration. Most theorems require that the smooth manifold be compact (which isn't true for my case). 
Let's just assume that the map is a fibration for now... what is its fiber? I'm pretty sure that it is $GL(k)$ since the action of $GL(k)$ on $M$ is free. If $A,B \in GL(k)$ and $\phi:\mathbb R^k \rightarrow V$ is a monomorphism then obviously $\phi A = \phi B$ implies that $A = B$ since $\phi$ is mono.
I will accept any answer that proves whether $M \rightarrow Gr(k,V)$ is a fibration/fiber bundle or not but I am also curious about what the general theory of actions of lie groups on manifolds has to say and if there are general theorems that my question is a special case of!
Thanks in advance for any help! 
 A: According to Moishe Kohan's answer here, if you have a free and proper action of a Lie group $G$ on a manifold $M$ (no assumptions about compactness on either $G$ or $M$), then the quotient map $M\rightarrow M/G$ is a $G$-principal bundle.
As you noted, there is a free action of $Gl(k)$ on $M$ given by precomposing.  So let's prove that this action is proper, meaning that the map $\phi:G\times M\rightarrow M\times M$ given by $\phi(g,m)\rightarrow (gm, m)$ is proper.  
To that end, first note the following, which confirms your guess that the fiber above any point is a copy of $Gl(k)$:
Proposition  Given two maps $f_1,f_2\in M$, there is a $g\in GL(k)$ with $f_1 \circ g = f_2$ iff the image of $f_1$ is the same as the image of $f_2$.  Further, if there is such a $g$, it is unique.
Proof:  First, since $g:\mathbb{R}^k\rightarrow \mathbb{R}^k$ is an isomorphism, it's surjective so $f_2 = f_1\circ g$ has the same image as $f_1$.
Conversely, if $f_2$ and $f_1$ have the same image, define $g$ as follows.  For each standard basis vector $e_i$ of $\mathbb{R}^k$, there is a unique element $v_i\in \mathbb{R}^k$ with $f_1(v_1) = f_2(e_i)$.  Let $g$ be the matrix whose $i$th column is $v_i$.  Then $f_1(g(e_i)) = f_1(v_i) = f_2(e_i)$, so $f_1\circ g = f_2$. 
Finally, if both $g, g'$ satisfy $f_2 = f_1 \circ g = f_1 \circ g'$, then $f_1 \circ g' g^{-1} = f_1$.  Since you've already noted the action is free, this foces $g'g^{-1} = I$, that is, $g' = g$.$\square$
We will also need the following lemma.
Lemma:  Suppose $f_i$ is a sequence in $M$ and $f_i\rightarrow f$.  Suppose $v_i$ is a sequence of vectors in $\mathbb{R}^k$ with $f_i v_i\rightarrow 0$.  Then $v_i\rightarrow 0$.
Proof:    Pick a background inner product on $V$ and use the usual inner product on $\mathbb{R}^k$.  Write each $v_i\in \mathbb{R}^k$ in polar form:  $v_i = r_i x_i$ with $x_i$ on the unit sphere $S^{k-1}$ in $\mathbb{R}^k$ and $r_i = |v_i|$.  Let us assume that $v_i$ does not converge to $0$, which means there is an $\epsilon \geq 0$ with the property that $r_i \geq \epsilon$ for some infinite set of $i$s.  Abusing notation, restrict to the subsequence of these $i$s and call the new sequence $v_i$.
Because $r_i f_i(x_i)  = f_i(v_i)\rightarrow 0$ and $r_i$ does is bounded below, we must have $f_i(x_i)\rightarrow 0$.  The $x_i$ all live on sphere, which is compact, so some subsequence (again called $x_i$) must converge to $x\in S{k-1}$.  Then $f_i(x) = f_i(x-x_i) + f_i(x_i)$.  Now, $\lim_{i\rightarrow\infty} f_i(x_i) = 0$ by hypothesis.  Further, $\lim_{i\rightarrow \infty} f_i(x-x_i) = 0$.  To see this, note that there is a universal bound $K$ for $\{f_i(v)|v\in S^{k-1}\}$ owing to the fact that $f_i\rightarrow f$.  Then $|f_i(x-x_i)|\leq K|x-x_i|\rightarrow 0$ as $i\rightarrow \infty$.
In short, $\lim_{i\rightarrow\infty} f_i(x) = 0$.  It now follows that $f(x) = 0$, which is a contradiciton because $f$ is injective and $x\in S^{k-1}$ so $x\neq 0$.  $\square$
Now, let's show the action is proper.  So, let $f_i$ and $g_i$ be sequences in $M$ and $G$ and assume that both $f_i\rightarrow f\in M$ and $h_i:=g_i\cdot f_i\rightarrow h\in N$.  We must show that some subsequence of the $g_i$ converges in $G$.
By the above proposition, the image of $f_i$ is the same as the image of $h_i$.  Again by the proposition, it follows that the image of $f$ and $h$ coincide, so there is a unique $g\in G$ with $h = f\circ g$.
We claim there is a subsequence of $g_i$ converging to $g$.  To that end, let us suppose for a contradiction that there is a neighborhood of $g$ which contains none of the $g_i$.  Thus, there is some $\epsilon > 0$ with the property that the matrix $g - g_i$ has at least one entry with is $\geq \epsilon$.  As there are only a finite number ($k^2$) of entries of a matrix in $Gl(k)$, there is at least one entry (say, in row $a$, column $b$), for which a subsequence of the $g_i$s (for which I'll abuse notation and call the subsequence $g_i$) satisfy $|(g-g_i)_{ab}|\geq \epsilon$.
Now, we know that $f_i \circ g_i \rightarrow f\circ g$.  So, $(f\circ g - f_i \circ g_i)(v)\rightarrow 0$ for any $v\in \mathbb{R}^k$.  Now, $$f\circ g - f_i \circ g_i = f\circ g - f_i\circ g + f_i\circ g - f_i\circ g_i = (f-f_i)\circ g + f_i\circ (g- g_i)$$ and $f_i\rightarrow f$.  This implies that $(f-f_i)(g v)\rightarrow 0$ for any $v\in \mathbb{R}^k$.  This, now, implies that $f_i\circ (g- g_i)(v)\rightarrow 0$ for any $v\in \mathbb{R}^k.$
However, let $v = e_a$ and set $w_i = (g-g_i)(e_a)\in \mathbb{R}^k$.  Then $|w_i|\geq \epsilon$ for every $i$, because the $b$-th entry of $w_i$ is, in absolute value, larger than or equal to $\epsilon$.  On the other hand, $f_i(w_i)\rightarrow 0$.  By the lemma, this implies $w_i\rightarrow 0$, which contradicts the fact that $|w_i|\geq \epsilon$.
Having reached a contradiction, we conclude that every open set about $g$ contains a $g_i$, so there is a subsequence of the original $g_i$ sequence converging to $g$.
