Is it possible that a path $a$ is not homotopic to the constant path? Is my argument correct?
Aim:
Let $X$ be a conected by paths set and $a:[0,1]\to X$ a path. So, $a\simeq c_a$, where $c_a(s)\equiv a(0)$ is the constant path based on $a(0)$.
I think this is trivial, because as $ X $ is connected, using a free homotopy $ H (a, t) = a_t $, where $ a_t (s) = a (ts) $ we deform $ a $ in order to obtain $ a \simeq c_a $.
Is it correct? Thank you so much.
edit $\simeq$ is the free homotopy
 A: It is correct. In fact, if $Y$ is any contractible space (as $I = [0,1]$) and $f  : Y \to X$ is any map to some space $X$ (no assumptions on $X$!), then $f$ is homotopic to a constant map. To see this, let $x_0 \in f(Y)$. Choose $y_0 \in Y$ such that $f(y_0) = x_0$. Since $Y$ is contractible, there exists a homotopy $H : Y \times I \to Y$ such that $H(y,0) = y$ and $H(y,1) = y_0$ for all $y \in Y$. Then $G = f \circ H : Y \times I \to X$ is a homotopy such that $G(y,0) = f(y)$ and $G(y,1) = x_0$ for all $y$.
Note that if we assume $X$ to be path connected, then $f$ is homotopic to each constant map $Y \to X$.
Let us emphasize that for $Y = I$ we cannot expect that the homotopy keeps fixed both endpoints $0, 1$ of $I$. This may confuse some readers because they misinterpret the word "homotopy" as "homotopy of paths" which by definition keeps endpoints fixed. However, here we deal with free homotopies. The best what can be achieved in general is that our homotopy keeps fixed a single point of $I$, for example $0$. Also note that that only closed paths have the chance that there exists a homotopy of paths to a constant path.
