Brian Bi
Exercise 2.1.1 Let $$a, b, c \in S$$. Then $$a(bc) = ab = a = ac = (ab)c$$ so the law of composition is associative. Obviously if $$|S| = 1$$, then the sole element of $$S$$ is the identity. Otherwise, suppose $$a \in S$$ and choose $$b \in S \setminus \{a\}$$; then $$ab = a \neq b$$, so $$a$$ is not the identity. Therefore there is no identity when $$|S| \neq 1$$.
1. Suppose $$a$$ has both a left inverse $$\ell$$ and a right inverse $$r$$, that is, $$\ell a = ar = 1$$. Then by associativity, $$\ell = \ell 1 = \ell(ar) = (\ell a)r = 1r = r$$. Therefore $$ar = 1$$ and $$ra = \ell a = 1$$, so $$r$$ is the inverse of $$a$$.
2. If $$a$$ is invertible, and $$b_1, b_2$$ are inverses of $$a$$, then part (a) implies that $$b_1 = b_2$$. Therefore inverses are unique for associative operations.
3. Observe that $$(ab)(b^{-1}a^{-1}) = a(b b^{-1})a^{-1} = a a^{-1} = 1$$, therefore $$b^{-1} a^{-1}$$ is the inverse of $$ab$$.
4. If $$S$$ is the set of functions $$\mathbb{N}^{\mathbb{N}}$$ with the law of composition given by the composition of functions, then the function $$f \in S$$ defined by $$f(x) = x+1$$ has a left inverse given by $$g(x) = \max(x-1, 0)$$ but no right inverse since there is no $$x$$ with $$f(x) = 0$$. Likewise the function $$g$$ has $$f$$ as its right inverse but can have no left inverse since it is not injective.
Exercise 2.1.3 This was explained in the solution to 1.2(d) above; we just note here that $$g(0)$$ can be any natural number, so there are infinitely many possible $$g$$'s.