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Let $N$ be a smooth $d$-dimensional connected orientable manifold $N$ 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 $f_t=tf_0+(1-t)f_1$.

I show below that a necessary condition is that $\pi_k(N)$ is trivial for every $1 \le k \le d$.

Question: Is this sufficient?

Edit: As showed by Qiaochu, if $\pi_k(N)$ are trivial for $1 \le k \le d$, then $N$ is contractible. So, does being contractible suffice?

(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.

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  • $\begingroup$ Was your stipulation that $M$ and $N$ both be $d$-dimensional deliberate? Because if so, then your comment that such manifolds must be simply connected because you can take $M=[0,1]$ only applies to $1$-dimensional manifolds. $\endgroup$
    – Jack Lee
    Commented Oct 15, 2017 at 18:02
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    $\begingroup$ You should also assume that $N$ is connected. Then it becomes a corollary of Whitehead's theorem. $\endgroup$ Commented Oct 18, 2017 at 12:22
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    $\begingroup$ Whitehead's theorem implies that $N$ is contractible, although not immediately: first observe that by the Hurewicz theorem if the first nontrivial homotopy group of $N$ occurs in degree $\ge n+1$ then so does the first nontrivial homology group, but since $N$ is $n$-dimensional all of its homology above degree $n$ vanishes, so $N$ has trivial homotopy groups. Next, as a manifold, $N$ has the homotopy type of a CW complex (surprisingly difficult), so Whitehead's theorem applies to any map to or from a point and shows that $N$ is contractible. Which I think is enough. $\endgroup$ Commented Oct 19, 2017 at 17:51
  • $\begingroup$ @QiaochuYuan Thanks! your explanation is great. Two more questions, if you please: (1) Is there an easy way to see that all the Homology groups of a manifold above its dimensions are trivial? (I mean an easier way than showing that the manifold is homotopy equivalent to a CW complex?) (2) So, now $N$ is contractible, as you say. Do you have an idea for why this implies the existence of a boundary respecting homotopy? $\endgroup$ Commented Oct 19, 2017 at 18:20
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    $\begingroup$ 1) It should follow from a suitable form of Poincare duality. 2) Morally speaking it's because the question, suitably asked, ought to have a homotopy-invariant answer, so one ought to be able to replace $N$ with a point. But I had some trouble writing out the details, and then you need to check that you can smooth homotopies. $\endgroup$ Commented Oct 19, 2017 at 18:54

1 Answer 1

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The proof is essentially the same as for manifolds without boundary:

  1. First, given any CW complex $X$ and a subcomplex $Y\subset X$ and a weakly contractible CW complex $Z$, every continuous map $f: Y\to Z$ extends to a continuous map $F: X\to Z$. This is a relative form of Whitehead's theorem and it is proven by induction on skeleta: You assume that the extension $F_k$ is constructed on the $k$-skeleton $X^k$ of $X$. For each $k+1$-cell $e$ of $X$, with the attaching map $\delta e$, if $e$ is in $Y$, you use $f$ to extend $F_k$, if $e$ is not in $Y$, then notice that the composition $F_k\circ \delta: S^k\to Z$ is null-homotopic, since $Z$ is weakly contractible. Hence, you obtain the required extension of $F_k$ to $e$ using this null-homotopy. You will find this in any algebraic topology book which covers the homotopy theory.

  2. You apply this to the manifold with corners $M'=M\times [0,1]$ (which admits a triangulation (the structure of a simplicial complex $X$) such that $Y=\partial M \times [0,1] \cup M\times \{0, 1\}$ is a subcomplex (use the fact that $M$ admits a triangulation such that $\partial M$ is a subcomplex). Also, $N$ admits a triangulation making it a simplicial complex $Z$. Now, consider the map $$ f: Y\to Z $$ which equals to $f_0, f_1$ on $M\times \{0\}, M\times \{1\}$ and which equals $f_0=f_1$ on $\partial M$. You obtain a continuous extension $F: M'\to N$. Now, if $M'$ was a smooth manifold with boundary, you could simply quote Whitney's extension theorem saying that if a smooth map defined on a submanifold admits a continuous extension then it also admits a smooth extension. You can find Whitney's theorem in many places, e.g. in Lee's book, "Introduction to Smooth Manifolds", 2nd edition, Chapter 6. In the case when the domain is a manifold with corners, you just need to observe that it embeds in a smooth manifold without boundary, so you use that as the domain of your map.

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