looking for a counterexample on bundles maybe anyone here could help me to understand some facts in bundles.
We have the following:


*

*Let $X$ be any space, and let $f:\underline{n}\rightarrow\underline{k}$ be a surjective map of bundles. Then $f$ has a splitting. ($\underline{n}$ means constant bundle with rank $n$).

*Let $X$ be a paracompact space. Then any surjection of bundle $E\twoheadrightarrow F$ has a splitting. Moreover, any injection of bundles $E \hookrightarrow F$ has a splitting.
So, as usual, with better assumptions on a base space, we have such a generalization in bundles level. 
Can anyone explain me or give an example of not paracompact space and such bundle map that has not splitting? Why the paracompactness is so important?
 A: On a paracompact space, every vector bundle admits a  bundle metric (a non degenerate positive form on each fiber which continuously depends on the fiber).
Using this metric every subbundle admits supplementary bundle (the orthogonal subbundle). 
In our case the kernel of the bundle map $  E\to F$ is a sub-bundle which admits therefore a supplementary bundle, or equivalently a splitting of the map $E\to F$. 
For the necessity of the "paracompacity" assumption to construct a metric, one should remember that if the tangent bundle admits a Riemanian metric, then the manifold is paracompact, just because the manifold itself is metrizable (the Riemanian distance does the job) and because every metric space is paracompact (Stone, Paracompactness and product spaces, Bull. Amer. Math. Soc. 54
(1948), 977-982)
Let $L$ be a long line (non paracompact), $T$ its tangent bundle. A splitting of $T\otimes T^* \to  \bf R$ (here $\bf R$ denotes the trivial bundle) would yield a non zero vector field, and therefore a Riemanin metric, contradicting the fact that the rela line is not metrizable. But such animals are not really interesting.
