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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .