# How could a path be homotopic to a point

I have been following Serge Lang's Intoduction to Complex Analysis at a Graduate Level and I met this theorem. I want to ask what does it mean for a function to be homotopic to a point? I am only familiar with paths being homotopic to each other but not to a point. Here is the extract

If anybody could shine some light onto this I will really appreciate it.

• "Point" here means "constant path at a point". Jul 15, 2020 at 22:35
• Hi @BenSteffan, by "constant path at a point" do you mean a path that consists of one point? Jul 15, 2020 at 22:38
• Yes, MathsWizard, that's precisely what it means. This is true when the set $U$ is simply connected. Jul 15, 2020 at 22:40
• Hi @TedShifrin, thank you for the comment. Could you clarify what you are referring to as being true for U simply connected? Jul 15, 2020 at 22:41
• In a simply connected region, every closed path is homotopic to a constant path. Most students encounter this notion, at least intuitively, in multivariable calculus. Jul 15, 2020 at 22:43

A path homotopy $$f\to g$$ is a homotopy $$H=H(x,t)$$ such that $$H(x,0)=f(x)$$ and $$H(x,1)=g(x).$$ A path $$f$$ is homotopic to a point $$c$$ if it is path homotopic to the constant path $$c(x)=c$$ for each $$x.$$
Consider this example: let $$f(x)=(\sin x,\cos x)$$ be the path which traces out the unit circle in $$\mathbb R^2$$. Then the path homotopy $$H(x,t)= tf(x)$$ has $$H(x,1)=f(x)$$ and $$H(x,0)=0$$ for each $$x.$$ It follows that $$f$$ is homotopic to the point $$0$$ since it is homotopic to the constant path at $$0.$$