Question about the proof that the Hilbert Cube is compact. Because of the fact that

$(1)$ The topological space $[0,1]$ is a continuous image of the Cantor space $(G,T)$.

There exists a mapping $\phi_n$ of $(G_n, T_n)$ onto $(I_n, T'_n)$ where, for each $n \in \Bbb N$, $(G_n, T_n)$ and $(I_n, T'_n)$ are homeomorphic to the Cantor space and $[0,1]$, respectively.
Additionally, there exists a continuous mapping $f : \prod (G_n,T_n)$ onto $\prod (I_n,T'_n) = I ^ {\infty}$.
The fact that 

$(2)$ If $(G_n, T_n)$ is a countable family of topological spaces each of which is homeomorphic to the Cantor space $(G,T)$, then $(G,T) \cong \prod_{i=1}^{\infty}(G_i, T_i) \cong \prod_{i=1}^{n}(G_i, T_i)$

Implies that $I^{\infty}$ is a continuous image of the compact space $(G,T)$, and hence is compact.
Question 1:  How is it that $(1)$ implies that the mapping $\phi_n$ exists?
Question 2:  I see that if $h : (G,T)$ homeomorphically to $\prod (G_i,T_i)$, then $ g = f \circ h: (G,T)$ continuously onto $I^{\infty}$, but how does this show that the continuous image of $(G,T) = I^{\infty}$?  Wouldn't $f$ have to be bijective in order for this to be true?
 A: Let $\varphi:G\to I$ be the continuous surjection whose existence is stated in $(1)$. For each $n\in\Bbb N$ let $\langle G_n,T_n\rangle=\langle G,T\rangle$, $\langle I_n,T_n'\rangle=\langle I,T'\rangle$, and $\varphi_n=\varphi$, thought of as a function from $G_n$ to $I_n$. If you take each $G_n$ to be homeomorphic to $G$ rather than identical to $G$, you have to work a little harder. For each $n\in\Bbb N$ let $h_n:G_n\to G$ be a homeomorphism, and let $\varphi_n=\varphi\circ h_n$; then $\varphi_n$ maps $G_n$ continuously onto $I$. And if in addition each $I_n$ is merely homeomorphic to $I$ rather than identical to $I$, for each $n\in\Bbb N$ let $k_n:I\to I_n$ be a homeomorphism, and define $\varphi_n=k_n\circ\varphi\circ h_n$: this maps $G_n$ continuously onto $I_n$.
As you say, $g=f\circ h:G\to I^\infty$ is a continuous surjection. That’s exactly what is meant by saying that $I^\infty$ is a continuous image of $G$: there is a continuous function mapping $G$ onto $I^\infty$. There is no requirement that the continuous function be a bijection. Indeed, the usual function $\varphi:G\to I$ used to map $G$ continuously onto $I$ is not a bijection: $\varphi\left(\frac13\right)=\varphi\left(\frac23\right)=\frac12$, for instance.
