# Slick proof that the inverse homotopy continuous?

While trying to learn algebraic topology from Hatcher, I frequently come across maps which are "obviously continuous" but for which it would seem tedious to actually show continuity by appealing to the open set definition.

For example, the text states "if $$f_0 \cong f_1$$ via $$f_t$$, then $$f_1\cong f_0$$ via the inverse homotopy $$f_{1-t}$$. At a conceptual level, my intuition is that reparametrizing a homotopy by $$t\mapsto 1-t$$ preserves continuity. I would like a proof that formalizes that if possible...but I don't know how to actually prove that the resulting inverse homotopy is continuous without looking at open sets in the image space and considering their preimages. (Specifically, define a homotopy $$G$$ such that $$G(s,t)=H(s,1-t)$$ where $$H(s,t)$$ is the homotopy from $$f_0$$ to $$f_1$$ and show preimages under $$G$$ of open sets are open).

• Presumably you mean "If $f_0\cong f_1,\dots$" but you write $f_0\cong f_0.$ Sep 10, 2023 at 17:43
• @ThomasAndrews thanks, edited Sep 10, 2023 at 17:48

This is actually straightforward. The open-set definition makes it easy to show the composition of continuous functions is continuous. The definition of product topology makes it easy to show the product of continuous functions is continuous. Using e.g. the metric space criterion for continuity, $$t\mapsto1-t$$ is evidently continuous (if $$|x-y|<\epsilon$$ then $$|(1-x)-(1-y)|=|x-y|<\epsilon$$ too).
So the composition: $$G:X\times I\overset{1\times(t\mapsto1-t)}{\longrightarrow}X\times I\overset{H}{\longrightarrow}Y$$Is continuous, as a composition of continuous functions.
• @Mithrandir No worries. In fact, this property of the product topology is arguably one of the main reasons we like it; if $f:A\to B$ and $g:A\to C$ are continuous, then $(f,g):A\to B\times C$ is continuous. This crucially holds for infinite products. See ‘universal property of the product’, if you can accept the abstraction Sep 10, 2023 at 17:55