Let $X = S^1 \times (0,1)$ be the cylinder. Every homeomorphism $f: X \to X$ is homotopic to either the identity $\operatorname{id}_X$, or to the map $i \times \operatorname{id}_{(0,1)}$ which inverts the orientation of $S^1$, i.e. $i:S^1 \to S^1, z\mapsto z^{-1}$.
Now suppose $f: X \to X$ is compatible with the projection $q: X \to (0,1)$, i.e. $q(f(x)) = q(x)$ for each $x \in X$. If $f$ is homotopic to $\operatorname{id}_X$, is it possible to find a homotopy $$H: [0,1] \times X \to X, \quad H_0 = f, \quad H_1 = \operatorname{id}_X$$ which is also compatible with $q$?
It turns out that this is rather simple. Call the two projections $$p: X \to S^1 \quad \text{and} \quad q: X \to (0,1).$$ By assumption, $q \circ f = q$, and we have a homotopy $$H: [0,1] \times X \to X, (t,x) \mapsto H_t(x)$$ with $$H_0 = f, \quad H_1 = \operatorname{id}_X.$$ Then define $$\tilde H_t(x) = (p(H_t(x)), q(x)).$$ This is clearly continuous and satisfies for all $x \in X$, $t \in [0,1]$: \begin{align} q(x) & = q(\tilde H_t(x))\\ \tilde H_0(x) & = (p(H_0(x)), q(x)) = (p(f(x), q(f(x)) = f(x) \\ \tilde H_1(x) & = (p(H_1(x)), q(x)) = (p(x), q(x)) = x. \end{align}