Is the smooth mapping space a deformation-retract of the continuous one?

Question

Let $X,Y$ be smooth manifolds.

Let $F \subset G$ denote the spaces of, respectively, smooth and continuous functions $X\rightarrow Y$; we give $F,G$ the compact-open topologies. (This should be compatible with the subspace topology on $F$ from $G$.)

Since every continuous map is homotopic to a smooth one by e.g. Whitney approxiomation, we know that $F$ meets every path-component of $G$.

I was wondering, is $F$ furthermore a deformation-retract of $G$?

The more natural question seems to me whether the inclusion is a (weak) homotopy equivalence. — Qiaochu Yuan, Jun 24 '22 at 21:23

freakish · Accepted Answer · 2022-06-24T20:42:21.720

6

No. For example if $X=Y=\mathbb{R}$ then the space of all smooth functions (even polynomials) is dense in $C(X,Y)$. And so it cannot be a retract of $C(X,Y)$. This idea generalizes to any smooth manifolds of positive dimension (in dimension $0$ smooth and continuous coincide).

edited Jun 24 '22 at 20:42

answered Jun 24 '22 at 20:27

freakish

47,446

Thank you, I hadn't considered the issue of density. – I.A.S. Tambe Jun 24 '22 at 20:30

Is the smooth mapping space a deformation-retract of the continuous one?

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