Are all vector bundles of the form $E=X\times V$ trivial? This might be a silly question, but bear with me.
Let $X$ be a topological space, $V$ a vector space and consider the vector bundle $E=X\times V$ with $\pi:E\to X$, $(p,v)\mapsto p$. This is the trivial $V$-bundle over $X$.
I had been under the impression that any vector bundle $E'=X\times V$ was trivial, but I realised that I had not been accounting for the fact that for a trivial bundle, $\pi$ must be projection onto the first factor. My question is, can we show that $E\cong E'$ in general, even when $\pi':E'\to X$ is not just projection onto the first factor?
To show this, I would need to find a homeomorphism $\phi:E\to E'$ such $\pi'\circ\phi=\pi$, and $\pi'=\pi\circ\phi^{-1}$. The obvious idea would be to use the local trivialisations of $E'$, but I don't quite see how to do so.
Any help or references would be much appreciated.
 A: As I mentioned in my comment above, the answer to the question is no, as demonstrated by the following example by Igor Belegradek in this MathOverflow answer which I have replicated below:

Consider the pullback $\xi$ of $TS^2$ via the projection of $S^2\times\mathbb R$ onto the first factor. The bundle $\xi$ is a nontrivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$ because its pullback under the inclusion $S^2\to S^2\times\mathbb R$ is $TS^2$, which is nontrivial. On the other hand, its total space is $\mathbb R\times TS^2$ which is diffeomorphic to $S^2\times\mathbb R^3$, which is the total space of the trivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$.

Let me add some details.
Let $\pi : TS^2 \to S^2$ be the projection, and $i : S^2 \to S^2\times\mathbb{R}$, $p \mapsto (p, 0)$ be the inclusion, so $\xi = i^*TS^2$. If $\operatorname{pr}_1 : S^2\times\mathbb{R} \to S^2$ denotes the natural projection map, then
$$\operatorname{pr}_1^*\xi = \operatorname{pr}_1^*i^*TS^2 = (i\circ\operatorname{pr}_1)^*TS^2 = \operatorname{id}_{S^2}^*TS^2 = TS^2.$$
As the pullback of a trivial bundle is trivial, and $\operatorname{pr}_1^*\xi = TS^2$ is non-trivial, we see that $\xi$ is a non-trivial vector bundle over $S^2\times\mathbb{R}$.
On the other hand, by definition, the total space of $\xi$ is given by
$$\xi = i^*TS^2 = \{((p, t), v) \in (S^2\times\mathbb{R})\times TS^2 \mid \pi(v) = p\}.$$
Now note that, as bundles over $S^2$, we also have
$$\varepsilon^1\oplus TS^2 = (S^2\times\mathbb{R})\oplus TS^2 = \{((p, t), v) \in (S^2\times\mathbb{R})\times TS^2 \mid \pi(v) = p\}$$
where the last equality is the definition of the direct sum of vector bundles.
As $TS^2 \to S^2$ is stably trivial, $\varepsilon^1\oplus TS^2 \cong \varepsilon^3 = S^2\times\mathbb{R}^3$. So the total space of the non-trivial rank two vector bundle $\xi \to S^2\times\mathbb{R}$ is diffeomorphic to $S^2\times\mathbb{R}^3$ and hence to the trivial rank two vector bundle over $S^2\times\mathbb{R}$, i.e. $(S^2\times\mathbb{R})\times\mathbb{R}^2$.
The point is that the total space of the non-trivial vector bundle $\xi \to S^2\times\mathbb{R}$ can also be viewed as the total space of a trivial bundle over $S^2$.
