I am working in the proof of Proposition 1B.9 of Hatcher's book, that sates a bijection $Hom(\pi_1(X),\pi_1(Y)) \simeq [X,Y]$ where $X$ is a CW-Complex and $Y$ is a $K(G,1)$. I have troubles to undestand the first construction.
Given a homomorphism $\phi: \pi_1(X,x) \to \pi_1(Y,y)$ we want to construct a continuos map $f:(X,x) \to (Y,y)$ such that $f_* = \phi$. The idea of the Hatcher proof is assume first that $X$ has a single $0$-cell (the base point $x$) and pick a map from the one skelleton $f:X_1 \to Y$. Because $X_1$ is a wedge sum of circles, each (close) $1$-cell $e_\alpha^1$ determines a element of $\pi_1(X,x)$. I think that the map that represents this element is given by the characterist map of $e_\alpha^1$ given by $\Phi_\alpha^1: D_\alpha^1 \to X$.
He let $f$ on the closure of $e_\alpha^1$ be a map representig $\phi([e_\alpha^1])$ and $i:X_1 \to X$ be a inclusion map, and he says that $\phi i_* = f_*$. I want to understand this.
First, $f$ on the closure means $e_\alpha^1$ is $f|_{\Phi_\alpha^1(D_\alpha^1)}$? Or $f$ (on the closure) is the composition $D_\alpha^1 \to \Phi_\alpha^1(D_\alpha^1) \simeq S^1 \to Y$ and them, $f$ (on the closure) is a loop in $Y$?
I'm not understand why $\phi i_* = f_*$. I have troubles to show that $\phi i_*[e_\alpha^1] = f_*[e_\alpha^1] $ because I am not undestand this construction.
If $e_\alpha^n$ is an $n$-cell with characteristic map $\Phi^n_\alpha\colon D^n_\alpha\to X$, then the closure of $e_\alpha^n$ is $\Phi^n_\alpha(D^n_\alpha)$. A map $f\colon X\to Y$ is continuous iff the restriction to the closure of each $n$-cell is continuous. In the case at hand $X_1$ is a wedge of circles and the closure of each $1$-cell is a wedge summand (i.e. a circle). If $e_\alpha^1$ is a $1$-cell, then the composite $$S^1\cong \bar{e}^1_\alpha\hookrightarrow X_1\hookrightarrow X$$ gives rise to an element in $\pi_1(X)$, which Hatcher denotes by $[e_\alpha^1]$. Now $\varphi([e_\alpha^1])\in \pi_1(Y)$. After choosing a representative of the homotopy class this yields a map $f_\alpha\colon \bar{e}_\alpha^1\cong S^1\xrightarrow{\varphi([e_\alpha^1])} Y$ and by the universal property of the wedge sum this yields $f\colon X_1\to Y$. On the higher dimensional skeleta everything is easier to define, since every map $S^{n-1}\to Y$ is nullhomotopic for $n\geq 2$.
As for your second question: The elements $[e^1_\alpha]$ generate $\pi_1(X)$ and $f_\ast([e_\alpha^1])$ is the homotopy class of the composite $$S^1\cong \bar{e}^1_\alpha\to X\xrightarrow{f}Y,$$ which by construction is equal to $$S^1\cong \bar{e}^1_\alpha\xrightarrow{f_\alpha}Y,$$ which itself by construction is equal to $\varphi([e_\alpha^1])$. Thus, $f_\ast([e_\alpha^1]) = \varphi([e_\alpha^1])$ and since the elements $[e^1_\alpha]$ generate $\pi_1(X)$ we see that $f_\ast = \varphi$.