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Constant maps $f_0, f_1 : X \rightarrow X$ are homotopic iff there is a continuous $F: I \rightarrow X$ with $F(0) = x_0$ and $F(1)=x_1$.

Forward direction $\Rightarrow$ If $G$ is the homotopy between $X \times I \rightarrow X$, then let $x'$ be fixed and define $F : I \rightarrow X$ by $F=f_t(x')$, then $f_0(x')=x_0$ and $f_1(x')=x_1$.

Opposite direction $\Leftarrow$ Here's where I'm stuck. I would say that just letting $f_t(x)$ be the homotopy should suffice since $f_0(x) = x_0$ and $f_1(x) = x_1$, but I feel like I might be assuming incorrectly by $f_t(x)$ being the homotopy.

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1 Answer 1

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I think your argument is right. To be more precise. Suppose there is a continuous map $F \colon I \to X$ such that $F(0)=x_0$, $F(1)=x_1$. Define $$G\colon I\times I\to X$$ given by $G(s,t)=F(t)$.

Then:

  • $G$ is continuous,
  • $G(x,0)=x_0=f_0(x)$ and
  • $G(x,1)=x_1=f_1(x)$.

So $G$ is the required homotopy.

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