# Do submersions induce open maps between spaces of differentiable maps?

Let $$X$$, $$Y$$ and $$Z$$ be smooth manifolds. Any differentiable map $$f \colon Y \rightarrow Z$$ induces a continuous map $$f_{\ast} \colon C^{\infty}(X, Y) \rightarrow C^{\infty}(X, Z)$$ via composition $$g \mapsto f \circ g$$, where the spaces of differentiable maps are equipped with the Whitney $$C^{\infty}$$ topology.

Question: When $$f$$ is a submersion, what can we say about the image of $$f_{\ast}$$ in $$C^{\infty}(X, Z)$$? Is it an open or dense or residual subset?

## 1 Answer

The density does not hold even in the compact connected case: if $$X=Y=Z=\mathbb S^1\simeq\mathbb R/\mathbb Z$$ and $$f:\theta\mapsto2\theta$$, then any composition $$f\circ g$$ has even degree, for instance it cannot be too close to the identity map.

I believe however that the following holds.

Proposition. If $$X$$ is compact, the composition operator $$f_*$$ is open in the Whitney topology.

In particular, it would imply that its range $$R$$ is open, so that for any pair of homotopic maps $$\phi\simeq\psi:X\to Z$$, $$\phi\in R$$ if and only if $$\psi\in R$$. In other words, and as the above example suggests, obstructions to being in $$R$$ are topological.

### Sketch of proof

I assume (without loss of generality, up to working component by component) that $$X$$, $$Y$$ and $$Z$$ are connected. I make the additional assumption that $$Y$$ is compact also, although I don't think it is needed; see below.

Proof. Let $$TY\to Y$$ (resp. $$TZ\to Z$$) be the the tangent bundle of $$Y$$ (resp. $$Z$$), and $$\ker Tf\to Y$$ be the kernel of the bundle map $$Tf:TY\to TZ$$. Note that $$\ker Tf$$ is indeed a vector bundle since $$f$$ is a submersion, and it is in fact a subbundle of $$TY$$. As such, it admits a complement, i.e. a subbundle $$F\to Y$$ of $$TY$$ such that $$F_y\oplus \ker T_yf=T_yY$$ at all points $$y$$ of $$Y$$.¹ In particular, $$T_yf$$ restricted to $$F_y$$ is an isomorphism for all $$y\in Y$$.

Identify a neighbourhood $$V$$ of the zero section in $$TY$$ to a neighbourhood $$V'$$ of the diagonal in $$Y\times Y$$, through a diffeomorphism $$\mu:V\to V'$$

Lemma. The function $$\alpha:(y,y')\in\mu(F\cap V)\mapsto(y,f(y'))\in Y\times Z$$ is a (smooth) diffeomorphism on a neighbourhood $$U$$ of the diagonal.

Assume for now that the lemma is true, and let's see how the proposition follows. Let $$f\circ g$$ be an element of the range $$R$$, and $$\phi_n:X\to Z$$ a sequence of maps converging to $$f\circ g$$ in the Whitney topology. We need to show that $$\phi_n=f\circ g_n$$ for all $$n$$ large enough, for a sequence $$g_n$$ that converges to $$g$$. The existence of $$g_n$$ is a consequence of the lemma: at some point, $$\phi_n$$ is sufficiently close to $$f\circ g$$ in the uniform topology so that $$(g(x),\phi_n(x))$$ belongs to $$\alpha(U)$$. Then, setting $$(g,g_n) := \alpha^{-1}\circ(g,\phi)$$, we have $$\phi_n=f\circ g_n$$ as announced. Now it suffices to show that $$g_n$$ converges to $$g$$ in the Whitney topology, i.e. that all their derivatives converge uniformly to that of $$g$$. But it is clear from the definition of $$g_n$$.

Proof of the lemma. The differential of $$\alpha$$ can be decomposed according to the structure of $$Y\times Y$$ and $$Y\times Z$$; along the diagonal, we have $$T\alpha=\begin{pmatrix}\mathrm{id}&*\\0& Tf_{|F}\end{pmatrix}.$$ According to the inverse function theorem, $$\alpha$$ is a local diffeomorphism on a neighbourhood of the diagonal. Then it is enough to show that it is injective on a smaller neighbourhood.

Let $$(y_n,y'_n)$$ and $$(\tilde y_n,\tilde y'_n)$$ be sequences with values in $$\mu(F\cap V)$$ such that they both approach the diagonal³ and $$\alpha(y_n,y'_n)=\alpha(\tilde y_n,\tilde y'_n)$$. It suffices to show that they coincide for an infinite number of $$n$$. In fact, $$y_n=\tilde y_n$$ by definition of $$\alpha$$. Up to considering a subsequence, we can assume that $$y_n$$ converges, say to $$y$$. Now if $$W$$ is a neighbourhood of $$(y,y)$$ in $$\mu(F\cap V)$$ over which $$\alpha$$ is a diffeomorphism, $$(y_n,y'_n)$$ and $$(\tilde y_n,\tilde y'_n)$$ must belong to that neighbourhood for $$n$$ large enough, so that they are in fact equal. This concludes the proof of the lemma.

### The non-compact case

We have used the fact that $$Y$$ is compact in the proof of the lemma. However, we need only assume that $$\alpha$$ is a diffeomorphism on a neighbourhood of the image of $$(g,g)$$, which is compact; the proof is then the same.

I don't know what happens in the case where $$Y$$ is compact but $$X$$ is not; I would guess it still holds, based on sheer intuition. If neither are, for $$f$$ the inclusion of an open arc in the circle, and $$g$$ the identity of said arc, $$f_*$$ is not open at $$g$$.

¹ For instance, fix a Riemannian metric on $$Y$$ and consider the orthonormal complement.

² For instance, $$\mu$$ could be the exponential map associated any Riemannian metric on $$Y$$.

³ In any reasonnable sense, say $$d(y_n,y'_n)\to0$$ for a Riemannian distance $$d$$.