Let $X$ be a geodesic metric space. Are there known local obstructions to the existence of a (bi-Lipschitz) homeomorphism between $X$ and $X\times \mathbb{R}$? 



Obstructions to being a trivial line bundle over itself
A homological obstruction that I mentioned in the comments can be extracted from  [On homotopical and homological $Z\_n$-sets](https://arxiv.org/abs/1002.2802)  by Taras Banakh, Robert Cauty, and Alex Karassev. 


This is a long paper but we won't need to know much. 
Given an abelian group $G$, a point $x\in X$ is a *$G$-homological $Z\_n$-set* if 
$H\_k(X, X-\{x\}; G)=0$ for all $k\le n$. The definition reduces to the one given on page 1 because $H\_k(X, X-\{x\}; G)=H\_k(U, U-\{x\}; G)$ for any open neighborhood $U$ of $x$ by excision in homology. For example, $0\in \mathbb R$ is a $Z\_0$-set but not a $Z\_1$-set.


Now suppose $G$ is $\mathbb Q$ or $\mathbb Z\_p$, and $X$ is a [Tychonov space](https://en.wikipedia.org/wiki/Tychonoff_space). 
Theorem 6.1 implies that if a point $x\in X$ is a $G$-homological $Z\_n$-set, then the point $(x,0)$ is a $G$-homological $Z\_{n+1}$-set in $X\times \mathbb R$. 


Therefore, if  **every** point of $X$ is a $Z\_n$-set, and a neighborhood of some $x\in X$ is homeomorphic to a neighborhood of $(y,t)\in X\times \mathbb R$, then $n=n+1$, and hence $n=\infty$. The latter is impossible if $x\in X$ is not a $Z\_{n+1}$-set, which gives an obstruction.


Note that no obstruction occurs when $X$ is the product of infinitely many copies of $\mathbb R$, and in fact, in this case $X$ is homeomorphic to $X\times \mathbb R$.


To apply the above one only needs to check whether the local homology is trivial (over $\mathbb Q$ or $\mathbb Z\_p$). Computing the local homology is not necessary.


