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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.

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    $\begingroup$ 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? $\endgroup$ Sep 23 '14 at 11:07
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    $\begingroup$ @DanieleZuddas: Indeed, this works just fine for smooth functions. I was mostly thinking about arbitrary sheaves where you don't have an addition function… $\endgroup$ Sep 23 '14 at 11:28
  • $\begingroup$ perhaps obstruction theory could be useful $\endgroup$ Sep 23 '14 at 13:29
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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.

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    $\begingroup$ This answer was given just a few minutes ago by @DanieleZuddas :) $\endgroup$
    – Yemon Choi
    Sep 23 '14 at 11:16
  • $\begingroup$ Indeed. I was concentrating on the case of arbitrary sheaves of sets (or ∞-groupoids), where there is no addition operation available, so I didn't try to use the additive structure… $\endgroup$ Sep 23 '14 at 11:32
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    $\begingroup$ I'm very sorry, when I started to write the answer the comment by @DanieleZuddas had not been posted. $\endgroup$ Sep 23 '14 at 11:39
  • $\begingroup$ That's fine, it wasn't meant as a criticism, more as a clarification for people who see this question later :) $\endgroup$
    – Yemon Choi
    Sep 23 '14 at 13:21

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