# Can a smooth function on a cross be extended to the whole plane?

Consider a real function on the union of two lines R×0 and 0×R in R² whose restrictions to R×0 and 0×R are smooth functions RR. Is it possible to extend this function to a smooth function on R²?

Motivation for this question comes from a desire to unstand better the notion of 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(R²×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.

• essentially you have two smooth functions $f_1, f_2 \colon \Bbb R \to \Bbb R$ such that $f_1(0) = f_2(0)$. Why don't just put $F(x, y) = f_1(x) + f_2(y) - f_1(0)$? Am I missing something? Sep 23, 2014 at 11:07
• @DanieleZuddas: Indeed, this works just fine for smooth functions. I was mostly thinking about arbitrary sheaves where you don't have an addition function… Sep 23, 2014 at 11:28
• perhaps obstruction theory could be useful Sep 23, 2014 at 13:29

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.