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. 1. 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? 2. 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. ![enter image description here](

1. 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 $x\in \\{a_1,a_2,\dots,\,d_1,d_2,\dots,\dots\\}$).
2. Yes, it is a typo.