How do I show that this map is path homotopic to a constant map? Let $\alpha:[0,1]\rightarrow \mathbb{C}\setminus\{0\}$ be null-homotopic loop.
Since $\mathbb{C}\setminus\{0\}$ is path connected, $\alpha:[0,1]\rightarrow \mathbb{C}\setminus\{0\}$ is homotopic to the constant map $\alpha(0)$.
It's intuitively clear that $\alpha$ and the constant map $\alpha(0)$ are indeed path-homotopic, but I dob't know how to prove this..
How do I prove this?
To angryavian:

 A: We can generalize slightly:
Proposition: Let $f$ be a loop in $X$ based at $x_0$. If $f$ is nullhomotopic, then there exists a homotopy from $f$ to the constant map at $x_0$ which is a path homotopy.
Proof: Let $f : I \to X$ be a loop based at $x_0$. 
Suppose $f$ is nullhomotopic, then there exists a homotopy $H : I \times I \to X$ such that $H(s, 0) = f(s)$ and $H(s, 1) = e_c(s) = c$ for some $c \in X$.
Define $\alpha_t : I \to X$ by $\alpha_t(s) = H(0, ts)$ and note that this is a path from $\alpha_t(0) = H(0, 0) = f(0) = x_0$ to $\alpha_t(s) = H(0, t)$.
Define $\beta_t : I \to X$ by $\beta_t(s) = H(1, t - ts)$ and note that this is a path from $\beta_t(0) = H(1, t)$ to $\beta_t(1) = H(1, 0) = f(1) = x_0$.
Define $h_t : I \to X$ by $h_t(s) = H(s, t)$ and note that this is a path from $h_t(0) = H(0, t)$ to $h_t(1) = H(1, t)$.
Hence, we've established that the following is well-defined: $\alpha_t * h_t * \beta_t$ is a loop based at $x_0$ for all $t \in I$.
If we define $F : I \times I \to X$ by $F(s, t) = (\alpha_t * h_t * \beta_t)(s)$, then it is a path homotopy between $f$ and the constant map at $x_0$.
