This is from Lemma 7.2, pg17
Let $E,F$ be vector bundles on $X$ and $f:E \rightarrow F$ a monomorphism on $Y$. Then if $\dim F > \dim E+ \dim X$, $f$ can be extended to a monomorphismon on $X$ and any two such extensions are homotopic rel $Y$.
I completely do not follow the supplied proof.
It would be great if someone could provide references (or proof ) for each of the claims.
Consider the fibre bundle $Mon(E,F)$ where the fibers are the monomorphisms of $E_x \rightarrow F_x$.
- This is homeomoprhic to $GL(n)/GL(n-m)$
- This space is $n-m-1$ connected.
- cross sections can be extended and are all homeomorphic if $$ \dim X \le \dim F- \dim E -1 $$
1) Let $V,W$ be $\mathbb{K}$-vector spaces with $\dim(V)\leq\dim (W)$. In particular there always there exists a linear monomorphism $\varphi:V\hookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $A\in Gl(W)$ then $A\varphi:V\hookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)\times Mon(V,W)\rightarrow Mon(V,W),\qquad (A,\varphi)\mapsto A \varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $\varphi_0\in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(\varphi_0)\cong Mon(V,W),$$
where $Stab(\varphi_0)\leq Gl(W)$ is the stabiliser subgroup of $\varphi_0$ with respect to the above action.
Now use the fixed $\varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^\perp\leq W$ to get a direct sum decomposition $V\oplus V^\perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^\perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^\perp)\hookrightarrow Gl(W)$ which sends $B$ to $id_V\oplus B$, and it is clear that
$$Stab(\varphi_0)=Stab(V)\cong Gl(V^\perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^\perp)\cong Gl(V\oplus V^\perp)/Gl(V^\perp)\cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $V\cong \mathbb{K}^n$ and $W\cong\mathbb{K}^{n+m}$ and take $\varphi_0$ to be the inclusion of $\mathbb{K}^n$ in $\mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(\mathbb{K}^{n+m})/Gl(\mathbb{K}^m)\cong Mon(\mathbb{K}^n,\mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $x\in X$, and so
$$Mon(E,F)\cong \bigcup_{x\in X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0\in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}\cong\mathbb{K}^n$ and $F_{x_0}\cong\mathbb{K}^{n+m}$ for some $n,m\in\mathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}\cong Gl(\mathbb{K}^{n+m})/Gl(\mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^{n+m})\rightarrow Mon(\mathbb{K}^n,\mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(\mathbb{K}^{n+m})$ is a Lie group, $Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$ is an open submanifold of $Mat_{n\times(n+m)}(\mathbb{K})$ and the action is smooth. Moreover $Gl(\mathbb{K}^n)$ is closed in $Gl(\mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(\mathbb{K}^n,\mathbb{K}^{n+1})$ is just the unit sphere
$$S(\mathbb{K}^{n+1})=\{x\in\mathbb{K}^{n+1}\mid |x|^2=1\},$$
and you can see this directly by thinking of points of $S(\mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $\mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^{n+1})\rightarrow S(\mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(\mathbb{K}^{n+1})=\{x\in\mathbb{K}^{n+1}\mid |x|^2=1\}=\begin{cases}S^n&\mathbb{K}=\mathbb{R}\\ S^{2n+1}&\mathbb{K}=\mathbb{C}\\ S^{4n+3}&\mathbb{K}=\mathbb{H}\end{cases}$$
so the map $Gl(\mathbb{K}^n)\rightarrow Gl(\mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(\mathbb{K}^{n+1})\rightarrow Gl(\mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(\mathbb{K}^n)\hookrightarrow Gl(\mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(\mathbb{K}^n)\hookrightarrow Gl(\mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(\mathbb{K}^n,\mathbb{K}^{n+m})$. If $E\rightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:X\rightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;\pi_kF)$, the first living in $H^{n+1}(X;\pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $\dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.