Prove that $r:\mathcal{Q} \to G(Y)$, $r(\alpha) = y_0$ when $y_0$ is such that $d(\alpha, G(Y)) = d(\alpha,y_0)$ is well defined Let $\mathcal{Q} = [0,1]^{\mathbb{N}}$, $Y= \lbrace \alpha=(\alpha_n)_n \in \mathcal{Q}:\alpha_1 = 0 \rbrace$ and $G(Y)= \lbrace tv + (1-t)\alpha: \alpha \in Y, t \in [0,1] \rbrace$ when $v=(1,0,0...)$ ($G(Y)$ is the geometric cone over $Y$). Prove that $r:\mathcal{Q} \to G(Y)$, $r(\alpha) = y_0$ when $y_0$ is such that $d(\alpha, G(Y)) = d(\alpha,y_0)$ is well defined.
I am trying to answer this question because I need to prove that $G(Y)$ is an Absolut Retract to $\mathcal{Q}$, i.e exist a continuous function $r:\mathcal{Q} \to G(Y)$  such that $r \circ r = Id_{\mathcal{Q}}$. The book [1] suggests that said function $r$ is as defined above. I just need the fact that $r$ is well defined.
To prove the existence of $y_0$ I did it as follows:
Since $G(Y)$ is a subcontinuum of $\mathcal{Q}$, $G(Y)$ is a compact subset of $\mathcal{Q}$. Let $\alpha \in \mathcal{Q}$. Since $d:\mathcal{Q} \times \mathcal{Q} \to (\mathbb{R}^{+} \cup \lbrace 0 \rbrace)$ is continuous and $\lbrace \alpha \rbrace \times G(Y)$ is a compact subset of $\mathcal{Q} \times \mathcal{Q}$ then exist $y_0 \in G(Y)$ such that $d(\alpha, G(Y)) = d(\alpha, y_0)$.
Since I could not prove the uniquenes of $y_0$ I try to define the function $r: \mathcal{Q} \to G(Y)$ where $r(\alpha) = r((\alpha_n)_n) = \alpha_1v + (1-\alpha_1)(0,\alpha_2, \alpha_3,...)$,  and so $r \circ r (Y) = Id_{Y}$. But it fails in the fact that $r \circ r (y) \neq y$ when $y \in G(Y)$ in general.
I appreciate the help of anyone who has an idea for this problem.
[1] Sam Nadler Jr. and Alejandro Wanes. HYPERSPACES: Fundamentals and Recent Advances. Pg 79, excercise 9.7.  
 A: For $\varepsilon \in [0,1]$ define a retraction $r_\varepsilon : [0,1] \to [0,\varepsilon]$ by $r_\varepsilon(t) = \varepsilon$ for $t \ge \varepsilon$ and $r_\varepsilon(t) = t$ for $t \le \varepsilon$.
Let us agree to write $\alpha =(\alpha_n) \in \mathcal Q$ in the form $\alpha = (t,\alpha_2,\alpha_3,\ldots)$.
Then define
$$r : \mathcal Q  \to \mathcal Q,  r(\alpha) = (t,r_{1-t}(\alpha_2), r_{1-t}(\alpha_3),\ldots) .$$
This is a continuous map. For $t = 1$ we have $r(\alpha) = v \in G(Y)$. Fot $t < 1$ let $\beta =  (0,r_{1-t}(\alpha_2), r_{1-t}(\alpha_3),\ldots)$ and $\alpha' = \beta/(1-t) \in Y$. Note that $\beta_n = r_{1-t}(\alpha_n) \in [0,1-t]$, thus $\beta_n/(1-t) \in [0,1]$. We see that $r(\alpha) = tv + (1-t)\alpha' \in G(Y)$. Thus $r(\mathcal Q) \subset G(Y)$.
Next let $\alpha = tv +(1-t)\alpha' \in G(Y)$ with $\alpha' = (0,\alpha'_2,\alpha'_3,\ldots)  \in Y$ and $t \in [0,1]$. Thus $\alpha = (t,(1-t)\alpha'_2,(1-t)\alpha'_3,\ldots)$ and $r(\alpha) = (t,r_{1-t}((1-t)\alpha'_2),r_{1-t}((1-t)\alpha'_3),\ldots) = (t,(1-t)\alpha'_2,(1-t)\alpha'_3,\ldots) = \alpha$ because $(1-t)\alpha'_n \in [0,1-t]$.
