; Q. Is there more polished form to write this procedure?

(let1 f (lambda (x y) (cons x y))  ; Actually, f is an arbitrary procedure.
  (map (lambda (outer-element)
         (map (lambda (inner-element)
                (f inner-element outer-element))
              '(a b c)))
       '(1 2 3)))

; ==> ((a . 1) (b . 1) (c . 1)
;      (a . 2) (b . 2) (c . 2)
;      (a . 3) (b . 3) (c . 3))