Show bijection between groups of homotopy classes I have to prove that $[X,S^1]_*$ is in a bijection with $[X,S^1]$, where $[...]_*$ is a pointed homotopy group with a distinguished point $x_o$ in $X$ and $1$ in $S^1$.
We have a hint to use for each $f : X \rightarrow S^1$ another function: $f^{'}$ such that $f^{'}(x) = f(x) * f(x_o)^{-1}$ and for each homotopy: $F : X \times I \rightarrow S^1$ a new homotopy $F^{'}(x,t) = F(x,t) * F(x_o,t)^{-1}$. But I don't know how to prove this bijection. Please, help me. 
 A: One more hint: you have an obvious inclusion of sets
$$
i: [X, S^1]_* \longrightarrow [X,S^1] \ .
$$
Haven't you? Ok, so the hint they gave to you says how to construct a map in the opposite direction:
$$
r : [X, S^1] \longrightarrow [X,S^1]_* \ .
$$
Namely, 
$$
r([f]) = [f'] \ ,
$$
where [f] denotes the homotopy class of $f$. 
I would try to prove that these two maps, $i,r$, are inverses one to another -besides being well-defined on homotopy classes.
For instance, without homotopy classes, it's already clear that $r\circ i = \mathrm{id}$, since, if $g: X \longrightarrow S^1$ preserves base points, then $g(x_0) = 1$. Hence $g'(x) = g(x)*g(x_0)^{-1} = g(x)$.
EDIT. You're right that the proof of $i\circ r = \mathrm{id}$ boils down to show that $f \sim f'$. But, as I said in my comment, the straight line homotopy doesn't work here because it's not contained in $S^1$. The second best idea is to normalize the straight line homotopy. That is, you quocient your $H(x,t)$ by its modulus $\vert H(x,t) \vert$, but in this case this doesn't work either (why?).
Instead, for every $x\in X$, you have to move from $f(x)$ to $f'(x)$ without exiting de circumference. How to do it? -Well, rotating from one point to another. Like this, for instance:
$$
F(x,t) = f(x)*e^{-it\theta}  \ .
$$
Here, $\theta$ is the argument of the complex number $f(x_0) \in S^1 \subset \mathbb{C}$ and
$$
e^{i\theta} = \cos\theta + i\sin\theta
$$
because of the Euler's formula. So:


*

*Who is $e^{-i\theta}$?

*Check that, indeed, this $F$ is the homotopy you were looking for.

