TITLE: Non trivial vector bundle over non-paracompact contractible space
QUESTION [23 upvotes]: The proof that the set of classes of vector bundles is homotopy invariant relies on the paracompactness and the Hausdorff property of the base space. Are there any known examples of:
Non trivial vector bundles over a paracompact non-Hausdorff contractible space
Non trivial vector bundles over a Hausdorff non-paracompact contractible space
Non trivial vector bundles over a non-Hausdorff non-paracompact contractible space

REPLY [3 votes]: To complement the answer providing a non-Hausdorff example, here is an example on a non-paracompact Hausdorff space as given in Schröer's Pathologies in cohomology of non-paracompact Hausdorff spaces arxiv:1309.2524.
I will slightly simplify the space in question, though.
Let $(X,0)$ be the wedge sum of countably many copies of intervals $([0,1],0)$. We denote the $1$ of the $n$-th copy by $1_n\in X$, $n\geq 1$. Of course, we have to alter the topology a little to get a non-paracompact space. Namely, the open subsets $U\subset X$ are those which are open in the CW-topology, and which either do not contain $0$, or if they do, they have to contain all but finitely many of the half-open intervals $[0,1_n)$.
This space is Hausdorff, as is easily seen, and contractible: each interval $[0,1_n]\subset X$ carries the usual topology and so $\{0\}\subset [0,1_n]$ is a strong deformation retract, which implies that each arm of $X$ can be contracted onto $0$ simultaneously.
Instead of an explicit example of a (complex) line bundle, Schröer shows that $H^1(X,U(1))\not = 0$, as follows.
Using that even for non-paracompact spaces, $H^1$-sheaf cohomology is computed by Čech-cohomology, it is first shown that $H^1(X,C^0(X))\not = 0$; then $H^1(X,U(1))\not = 0$ follows from the real Euler sequence $0\to\underline{\mathbb Z}_X\to C^0(X)\xrightarrow{exp(\cdot 2\pi i)}\underline{U(1)}_X\to0$, and $H^1(X,\underline{\mathbb Z}_X) = 0$.
Instead of giving all the details, let me directly construct a non-trivial real line bundle, but using the same idea.
Consider the open cover of $X$ defined by $U_0 = \bigcup_{n\geq 1}[0,1_n)$ and $U_n = (0,1_n]\subset X$, $n\geq 1$.
Then clearly $U_n\cap U_m=\emptyset$ unless $n = m$ or $n = 0$ or $m=0$.
Hence, we have to specify transition functions $g_n\colon U_0\cap U_n = (0,1_n)\to\mathbb{R}^\times$, $n\geq 1$.
Let $f_n\colon U_n\to\mathbb{R}$, $n\geq 1$, be continuous functions with $f_n(1) = 0$, but $f_n(x)\not=0$ for $x\not=1$, and let $g_n = \tfrac{1}{f_n}$.
(E.g., $g_n(x) = (1-x)^{-1}$.)
Let $L$ be the line bundle defined by those transition functions.
I claim that every section $s\colon X\to L$ is trivial at all but finitely many $1_n$.
In particular, $L$ itself is non-trivial.
Let $s\colon X\to L$ be any section. It defines (and is defined by) continuous functions $s_n\colon U_n\to \mathbb{R}$, $n\geq 0$, satisfying $s_n(x) = f_n(x)s_0(x)$ for all $x\in (0,1_n)$, $n\geq 1$.
From the existence of the limit $\lim_{x\to 1}f_n(x) = f_n(1) = 0$, we conclude that $s_n(1_n) = 0$ as soon as the limit $\lim_{x\to 1_n}s_0(x)$ exists as well.
To see that this happens for all but finitely many $n$, it suffices to observe that $s_0$ is bounded away from finitely many $U_n$, which is a consequence of our choice of topology.