Prove that these loops are homotopic

Question

Let $G$ be a topological group with identity element $e$. Let $f,g: (S^1, (1,0)) \to (G,e)$ be loops in $G$ with base point $e$. We define $f * g: (S^1, (1,0)) \to (G,e)$ by $$f * g(s) = f(t) \cdot g(t)$$ where $\cdot$ is the group multiplication. Then $fg$ and$ f*g$ are homotopic.

[Note: $fg$ denotes the "concatenation" of $f$ and $g$, i.e. we first walk the $f$ and then $g$, both at double speed.]

So we need to find a homotopy from $fg$ to $f*g$. I've tried writing down various possible homotopies but none of them worked and I can't seem to figure it out. I appreciate any hint.

Balarka Sen · Accepted Answer · 2014-12-16T20:45:15.440

6

Hint : Consider $(f e) * (e g)$. This is obviously homotopic to $f * g$. Can you also show that this is homotopic to $fg$?

Edit : Here's the whole solution, posted on request from the OP.

By definition of path concatenations, $fe : [0, 1] \to G$ is the path $$fe(t) = \begin{cases} f(2t) & 0 \leq t \leq \tfrac{1}{2} \\ e(2t-1) = e & \tfrac{1}{2} \leq t \leq 1 \end{cases}$$ And $eg : [0, 1] \to G$ is the path $$eg(t) = \begin{cases} e(2t) = e & 0 \leq t \leq \tfrac{1}{2} \\ g(2t-1) & \tfrac{1}{2} \leq t \leq 1 \end{cases}$$

Thus, $$(fe) * (eg) = \begin{cases} f(2t) * e & 0 \leq t \leq \tfrac{1}{2} \\ e * g(2t-1) & \tfrac{1}{2} \leq t \leq 1 \end{cases} \\ \;\;\;\;\;\;\;\;\;\; = \begin{cases} f(2t) & 0 \leq t \leq \tfrac{1}{2} \\ g(2t-1) & \tfrac{1}{2} \leq t \leq 1 \end{cases}$$

Which is just the path $fg$. Hence, $f * g \sim (fe) * (eg) \sim fg$, and the homotopy follows.

edited Dec 16 '14 at 20:45

answered Dec 16 '14 at 16:11

Balarka Sen

14,380

I know that the answer was given long time ago, but can I ask you, please. – Kerr Dec 15 '15 at 03:29
@Jane Sure, ask away what you want to ask. – Balarka Sen Dec 15 '15 at 03:30
do you use the path-connectedness of $G$ here? – Kerr Dec 15 '15 at 03:31
@Jane Yes, of course. – Balarka Sen Dec 15 '15 at 03:37
but if $f$ and $g$ are loops and $e$ is constant map, why do we need path connectedness? – Kerr Dec 15 '15 at 03:39
@Jane To homotope stuff. Of course, you can do these in path-components, but you need to be careful. – Balarka Sen Dec 15 '15 at 03:41
ah, you mean the path-connectedness of $G$ guarantee the homotopy between $f*g$ and $fg$? – Kerr Dec 15 '15 at 03:43
@Jane Somewhat. But as I said, you can do all of these in the path compoenent of $e$ in $G$. – Balarka Sen Dec 15 '15 at 03:44
I see now, thank you very much! – Kerr Dec 15 '15 at 03:47

Prove that these loops are homotopic

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