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.