This is a cross-post from MSE.

Let $N$ be a smooth $d$-dimensional connected orientable manifold which have the following property:

For every smooth $d$-dimensional manifold $M$ with non-empty boundary, and for every smooth maps $f_0,f_1:M \to N$ such that $f_0|_{\partial M}=f_1|_{\partial M}$, there exist a (smooth) homotopy $f_t$ which respects the boundary, i.e such that $f_0|_{\partial M}=f_t|_{\partial M}$ for all $t$.

(Note I am only testing $N$ with "sources" $M$ of the same dimension.)

$N=\mathbb{R}^d$ is an example; take $f_t=tf_0+(1-t)f_1$.

A necessary condition for $N$ to have this property is $\pi_k(N)=\{1\}$ for $1 \le k \le d$. (See below if interested). One can show then that the theorems of Whitehead and Hurewicz imply $N$ is contractible.

Question: If $N$ is contractible, does it have the property?

(For a start, let's try to see if there exists a continuous boundary respecting homotopy, and worry later about smoothing it).

Proof that $\pi_k(N)=\{1\}$ is necessary:

Suppose $N$ has the property, and let $\alpha_1,\alpha_2:(\mathbb{S}^k,p) \to (N,q)$. Since $\mathbb{S}^k \cong D^k /\partial D^k$ We can think of the $\alpha_i$ as maps $D^k \to N$ taking the boundary $\partial D^k$ to $q$.

Let $M=D^k \times \mathbb{R}^{d-k}$, and define $f_i:M \to N$ by $$ f_i(t,x)=\alpha_i(t).$$

Then $f_0|_{\partial M}=f_1|_{\partial M}$. By assumption, there exist a boundary respecting homotopy $f_s$;

Now $f_s(\cdot,0)$ is a homotopy of $\alpha_1,\alpha_2$ fixing the boundary.

  • $\begingroup$ OK well in the meantime you may be amused to read this which is an advanced case of a popular delusion. $\endgroup$ – Mikhail Katz Nov 2 '17 at 13:09

Let me first talk only about continuous maps. Your question becomes equivalent to asking whether every map from $\partial (M\times I) \to N$ (for $I$ the interval) can be extended to $M\times I$. This follows from the following two facts.

1) $\partial (M\times I) \to M\times I$ is cofibration.

2) If $i: A \to B$ is a cofibration, $X$ is any space and $f: A \to X$ and $g: B \to X$ are maps such that $gi \simeq f$, then there exists a $g': B \to X$ such that $g'i = f$.

We apply this to the case $A = \partial (M\times I)$, $B = M\times I$ and take $g$ to be a constant map; as $N$ is contractible the corresponding triangle commutes up to homotopy and we can use (2).

Proof of 1): The inclusion $\partial (M\times I) \to M\times I$ is a neighborhood deformation retract (NDR) since it has a collar. (This is true for every topological manifold with boundary by Brown's Locally flat imbeddings of topological manifolds, but there is of course a simpler argument available in our case.) Every NDR is a cofibration.

Proof of 2): Let $H: A \times I \to X$ be a homotopy with $H|_{A\times 0} = gi$ and $H|_{A\times 1} = f$. Because $i$ is a cofibration, we can extend this to a map $H': B \times I$ that agrees with $H$ on $A\times I$. We set $g' = H'|_{B\times 1}$.

To obtain a smooth homotopy, one should first make sure that our map $M\times I \to N$ is smooth on a neighborhood of the boundary using a collar; then one can apply a smooth approximation theorem that does not change the map on a given closed subset such that the map was already smooth in a neighborhood of it.

  • $\begingroup$ Thanks. Is $\partial (M\times I)=M \times \partial I=M \times \{0\} \cup M \times \{1\} \cup \partial M \times [0,1]$? (Is this the standard topological boundary?) $\endgroup$ – Asaf Shachar Oct 24 '17 at 4:19
  • $\begingroup$ If you delete the $M\times \partial I$ in the middle, your equation is correct; and yes, it agrees with the standard topological boundary. $\endgroup$ – Lennart Meier Oct 24 '17 at 6:36

The answer to your question is yes. Let me sketch a proof.

NOTE: In the following I'm using simplicial sets because I don't want to mess with the topologies in the space of smooth maps. Someone more geometrically minded than myself can probably easily adapt the proof to use spaces of smooth maps rather than simplicial sets.

If $M, N$ are two smooth manifolds (possibly with boundary) I denote by $\mathrm{Map}(M,N)$ the simplicial set whose $n$-simplices are the smooth maps $$ M\times\Delta^n\to N\,.$$ Note that $\mathrm{Map}(M,N)$ is automatically a Kan complex.

Lemma: For any manifold $M$ (possibly with boundary) the space of smooth maps $\mathrm{Map}(M,N)$ is contractible.

Proof: Since $N$ is contractible, let us choose a smooth homotopy $h:N\times \Delta^1\to N$ of the identity with the constant map at $x\in N$. Then the map

$$\mathrm{Map}(M,N)\times \Delta^1\to \mathrm{Map}(M,N)$$ sending a pair $$(f:M\times \Delta^n\to N, g:\Delta^n\to \Delta^1)$$ to $$M\times \Delta^n \xrightarrow{(f,g\circ\mathrm{pr}_2)}N\times \Delta^1\xrightarrow{h} N$$ is a homotopy of the identity on $\mathrm{Map}(M,N)$ with the constant map at the constant map at $x$. $\square$

Lemma: If $M$ is a smooth manifold with boundary $\partial M$, the map $$\mathrm{Map}(M,N)\to \mathrm{Map}(\partial M,N)$$ is a Kan fibration.

Proof: Using a collar for $\partial M$, you can prove that the inclusion $M\times |\Lambda^n_i|\cup \partial M\times \Delta^n \subseteq M\times \Delta^n$ has a smooth retraction. $\square$

Finally, the result you want is equivalent to the existence of a diagonal arrow in the following diagram

$$ \require{AMScd} \begin{CD} \partial \Delta^1 @>>> \mathrm{Map}(M,N)\\ @VVV @VVV\\ \Delta^1 @>>> \mathrm{Map}(\partial M,N)\\ \end{CD} $$

where the top arrow picks your two maps $f_0,f_1$ and the bottom arrow is a smooth homotopy of their restriction to the boundary.

But the left vertical map is an inclusion of simplicial sets and the right vertical map is a trivial Kan fibration (being a Kan fibration between contractible simplicial sets), so the diagonal arrow always exists.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.