Let X be a topological space, $Y\subset \mathbb{R}^n$ be convex, then any continuous map $ f\in\mathcal{C}(X,Y) $ is null-homotopic.
We can write the homotopy from $ f$ to constant map explicitly, i.e. $$F:[0,1]\times X\to Y, t\mapsto ty_0+(1-t)f(x),$$ where $y_0\in Y$ is arbitrary.
But it is a little difficult for me to show $ F $ is continuous rigorously. Actually, what we really need to prove is that $$G:[0,1]\times Y\to Y,t\mapsto ty_0+(1-t)y$$ is continuous, can anyone help me?
Note that $[0,1]\times Y$ is metrizable, and so it suffices to prove that if $(t_n, y_n)$ is a sequence converging to $(t, y),$ then $G(t_n, y_n)$ converges to $G(t, y).$ This is fortunately not so hard.
Note that, since addition is continuous, it suffices to verify that $t_ny_0 \rightarrow ty_0$ and $(1-t_n)y_n \rightarrow (1-t)y.$
The first identity is easy, since it is equivalent to $(t_n-t)y_0 \rightarrow 0.$ As $y_0$ is fixed, and $|t_n - t| \rightarrow 0,$ we get the desired limit.
The second identity requires some work. First, split it up as $y_n - t_ny_n \rightarrow y - ty.$ Using continuity of subtraction, it suffices to prove $y_n\rightarrow y$ (which we already know is true) and $t_ny_n\rightarrow ty.$ This second claim is similar to the product rule; it can be proven by noticing that
$$||t_ny_n - ty|| \leq ||t_ny_n - t_ny|| + ||t_ny - ty|| \leq |t_n| \cdot ||y_n - y|| + |t_n-t| \cdot ||y||.$$
Since the sequence $t_n$ is bounded above by 1, and since $||y||$ is fixed, as $n\rightarrow\infty,$ the right side of the above goes to zero, so that we have the desired limit.