- #1
littleHilbert
- 56
- 0
Hello,
Here is a short lemma:
A path-connected space X is simply-connected iff any two loops in X are free homotopic.
My question is whether it is allowed to use a straight-line homotopy straight away in order to construct a free homotopy? For example, let u and v be two loops and w is a curve from point a to point b. Then:[itex]
\begin{equation} H_f(t,s):=
\begin{cases} (1-3s)u(t)+3sa,\ \text{for $0\leqslant{}s\leqslant\frac{1}{3}$}; \\
w(3s-1),\ \text{for $\frac{1}{3}\leqslant{}s\leqslant\frac{2}{3}$}; \\
(3-3s)b+(3s-2)v(t),\ \text{for $\frac{2}{3}\leqslant{}s\leqslant1$}.
\end{cases}
\end{equation}
[/itex]
(there is an error in the latex-output: "0" instead of "sh")
That would actually do, wouldn't it? I mean a Nullhomotopy is not necesserally a straight-line homotopy. So maybe it's a loss of generality?
Here is a short lemma:
A path-connected space X is simply-connected iff any two loops in X are free homotopic.
My question is whether it is allowed to use a straight-line homotopy straight away in order to construct a free homotopy? For example, let u and v be two loops and w is a curve from point a to point b. Then:[itex]
\begin{equation} H_f(t,s):=
\begin{cases} (1-3s)u(t)+3sa,\ \text{for $0\leqslant{}s\leqslant\frac{1}{3}$}; \\
w(3s-1),\ \text{for $\frac{1}{3}\leqslant{}s\leqslant\frac{2}{3}$}; \\
(3-3s)b+(3s-2)v(t),\ \text{for $\frac{2}{3}\leqslant{}s\leqslant1$}.
\end{cases}
\end{equation}
[/itex]
(there is an error in the latex-output: "0" instead of "sh")
That would actually do, wouldn't it? I mean a Nullhomotopy is not necesserally a straight-line homotopy. So maybe it's a loss of generality?