Q:

Clarification on the proof of theorem 8 (Set Theory and Metric Spaces - Kaplansky)

There are 2 parts (red underline and yellow highlight) in the proof below that I couldn’t follow. 

Devise a one-to-one map of A onto C. Doesn’t the article ‘a’ mean ‘one’? However, the proof says: take g = the identity (for set D) AND take g = f (for unilateral cycles E). That means two instead of one function g, doesn’t it?
Is it a typo for the highlighted phrase (C might adjoin to A)? It should be C might adjoin to f(A) instead, shouldn’t it?

Could someone please help clarify? Thank you in advance. 

A:

Yes, we are constructing one function on $A$, but separately on $D$ and (two parts of) $E$.
Starting from the given example, $f(A)=D\cup\{a_2,a_3,\dots\}\cup\{b_2,b_3,\dots\}\cup \dots$, assume specifically that $C=f(A)\cup\{b_1,c_1\}$.
Then set $g(x)=x$ if $x\in D$, 
$g(x)=x$ if $x\in \{b_1,b_2,\dots,\,c_1,c_2,\dots\}$, 
and $g(x)=f(x)$ otherwise (so that if $x\in \{a_1,a_2,\dots,\,d_1,d_2,\dots,\dots\}$).
Yes, it is a typo. 

