Homotopy: equivalence relation (continuity of homotopy in symmetry)

Question

Let $f_0, f_1: X \rightarrow Y$ be continous on topological spaces $X,Y$. Let $F$ be a homotopy between $f_0, f_1$. Which argument shows that $G(x,t):=F(x,1-t)$ is continous?

Seems to be hard to deduce this considering preimages of open sets. I hope that one can say "G is continous in t for every x and also continous in x for every t, thus continous", but what would be the foundation for this argument (however, this argument is probably wrong)? (It's not the universal property of the product, since we're considering a function whose domain is a product and not the image)

Answer 1

• $r(t) : t \mapsto 1-t$ is a continuous function $[0,1] \to [0,1]$
• $(1_X, r)$ is thus a continuous function $X \times [0,1] \to X \times [0,1]$
• $F \circ (1_X, r)$ is thus a continuous function $X \times [0,1] \to X$

Answer 2

As the user Nobody pointed out, the map $F$ is continuous as it is a composition of continuous maps:

The map $\iota: [0,1] \rightarrow [0,1]$, $t \mapsto 1-t$, and the Homotopy $G$. Then the map $F = G \circ (Id_X , \iota)$, is continuous as both $G$ and $\iota$ are contiuous.