If $f\circ q,g\circ q$ are homotopic on $S^1$, are $f,g$ homotopic? Let $f,g:S^1\rightarrow S^1$ be continuous functions.
Define $\alpha:[0,1]\rightarrow S^1:t\mapsto (\cos 2\pi t, \sin 2\pi t)$.
If $f\circ \alpha$ and $g\circ \alpha$ are homotopic, then are $f,g$ homotopic?
How do I prove this?
 A: Notice that since $I$ is contractible and $S^1$ is connected, $f \circ \alpha, g \circ \alpha : I \to S^1$ are always going to be homotopic. A proof of this statement is included below.
However, if your statement is correct, it would follow that $f, g : S^1 \to S^1$ are always homotopic, which isn't true.

A space $X$ is contractible if the identity map $i_X : X \to X$ is nullhomotopic.
Lemma: If $X$ is contractible and $Y$ is path connected, then $f, g : X \to Y$ are homotopic.
Proof: Since $X$ is contractible, $i_X \simeq e_c$ for some $c \in X$. 
So there exists a homotopy $H : X \times I \to X$ with $H(x, 0) = e_X(x) = x$ and $H(x, 1) = e_c(x) = c$.
Notice that $f \circ H$ is a homotopy between $f = f \circ i_X$ and $f \circ e_c = e_{f(c)}$. and $g \circ H$ is a homotopy between $g = g \circ i_X$ and $g \circ e_c = e_{g(c)}$. 
Since $Y$ is path connected, there exists a path $\alpha : I \to Y$ with $\alpha(0) = f(c)$ and $\alpha(1) = g(c)$. This induces a homotopy between $e_{f(c)}$ and $e_{g(c)}$: Define $K : Y \times I \to Y$ by $K(s, t) = \alpha(t)$; observe that $K(s, 0) = \alpha(0) = f(c)$ and $K(s, 1) = \alpha(1) = g(c)$.
Hence we have $f \simeq e_{f(c)} \simeq e_{g(c)} \simeq g$.
Conclude that $f$ is homotopic to $g$.
