Consider a real function on the union of two lines **R**×0 and 0×**R** in **R**² whose restrictions to **R**×0 and 0×**R** are smooth functions **R**→**R**.
Is it possible to extend this function to a smooth function on **R**²?


Motivation for this question comes from a desire to unstand better
the notion of [concordance](http://ncatlab.org/nlab/show/concordance), in particular, how concordances can be composed.
Recall that a concordance of two sections x,y∈F(X) of some sheaf (or ∞-sheaf) F on the site of smooth manifolds is a section A∈F(**R**×X) whose restrictions along the inclusions 0×X→**R**×X and 1×X→**R**×X are x and y respectively.
Given a concordance A from x to y and a concordance B from y to z one can
ask if they can be composed to a concordance C from x to z.
(Such a composition is necessarily nonunique.)
One way to do this is to find a section v∈F(**R**²×X) whose restrictions to 0×**R**×X and **R**×0×X are A respectively B, where x, y, and z live over the points (1,0), (0,0), and (0,1).
Then restricting to the line passing through points (1,0) and (0,1) gives a concordance from x to z.
The above question then asks if it is possible to compose concordances in such a way for the sheaf of smooth functions, i.e., F(X)=C^∞(X).
Needless to say, there are other ways to compose concordances,
for example, using sitting instances, but I would like to know
if this particular method makes sense at least in some cases.



Can a smooth function on a cross be extended to the whole plane?
It is possible, indeed.


Assume that you have smooth functions $f:\mathbb{R} \longrightarrow \mathbb{R}^2$ and $g: \mathbb{R} \longrightarrow \mathbb{R}^2$ such that $f(0)=g(0)$. Define $h:\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ by $h(x,y) = f(x) + g(y) - f(0) $. Then, $h$ is clearly smooth and satisfies $h(x,0) = f(x) + g(0) - g(0) = f(x)$ and $h(0,y) = f(0) + g(y) - f(0) = g(y)$, as wanted.


