Are homotopic maps with identical boundary values homotopic through a boundary preserving homotopy?

Question

Let $M,N$ be smooth $d$ dimensional manifolds. Suppose $f_0,f_1:M \to N$ are homotopic, and that $f_0|_{\partial M}=f_1|_{\partial M}$.

Is there a homotopy $f_t$ such that $f_0|_{\partial M}=f_t|_{\partial M}$ for all $t$?

Answer 1

Not necessarily.

As a counterexample, take $M,N=S^1\times[0,1],\;f_0=id,\;f_1(\theta,t)=(\theta+2\pi t,t)$.

Edit: We show that there is no homotopy between the above $f_0,f_1,$ that restricts to the identity on the boundary at any time. Define two paths $\gamma_0,\gamma_1:[0,1]\to S^1\times[0,1]$ by $$ \gamma_0(t)=(0,t),\quad\gamma_1(t)=(2\pi t,t). $$ If the above $f_0,f_1,$ are homotopic without moving the boundary then, in particular, the paths $\gamma_0,\gamma_1$ can be extended to a smooth family of paths $\gamma_t:[0,1]\to S^1\times[0,1]$ such that $\gamma_t(0)=(0,0)$ and $\gamma_t(1)=(0,1)$ for every $t$. Now, for every $t$, write $$ I(t):=\int_{\gamma_t}d\theta. $$ Then $I(t)$ changes continuously with $t$ and only admits values in $2\pi\mathbb{Z}$. It follows that $I(0)=I(1)$, which is a contradiction.

Remark: This example is the first step in the definition of Dehn twists. Hence, I'm pretty sure you can find more explanations and intuitions for that across the internet or in any textbook about mapping class groups.