Suppose $f , g : X \to U \subset \mathbb R^2$ are two mappings from a topological space $X$ to a convex set $U$.
Prove that $f$ and $g$ are homotopic, using only the definition of the product mapping.
Here, I get the homotopy to be $(1-t)f(x) + t g(x)$, but I am not able to prove the continuity of this mapping.
On
@Scar,
Notice that the convex set $U \subset \mathbb{R}^2$ is contractible. This means that there is only one element in the homotopy equivalence class $[X,U]$.
To prove this, let $f,g \in [X,U]$. Then, by contractibility of $U$, there exists some homotopy $H: U \times I \to U$ that is the homotopy from the identity map in $U$ to a constant loop in $U$. But now we can define a homotopy $F: X \times I \to U$ defined by $F(x,t) = H(f(x), 2t), \space t \in [0, 1/2]$ and $F(x,t) = H(g(x), 2-2t), \space t \in [1/2, 1]$. Notice that $F(x, t) = H(f(x),1) = c = H(g(x), 1)$ at $t = [1/2]$, where $c$ is the constant loop in $U$. So $F$ is continuous by the Pasting Lemma, and defines a homotopy between the maps $f,g : X \to U$
The function $$H:[0,1]\times \mathbb{R}^2 \to\mathbb{R}^2 $$ defined $$f(t,u.v)= (1-t) u +tv$$ is continuous and hence the composition $$H(t, f(x) ,g(x))$$ is also continuous.