a problem on composition of functions

What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.

Suppose $f$ is not surjective, and let

$B = f(A) \tag 1$

be the image of $A$ under $f$; since

$f \circ f = f, \tag 2$

it is clear that every element of $B$ is fixed under $f$, for

$b \in B \Longrightarrow \exists c \in A, \; b = f(c) \Longrightarrow f(b) = f(f(c)) = f(c) = b; \tag 3$

furthermore, for $c \in A$,

$f(c) = c \Longrightarrow c \in B; \tag 4$

thus $B$ is precisely the set of fixed points of $f$.

We have assumed $f$ not surjective; then by the above we have

$B \subsetneq A, \tag 5$

which implies

$\exists a \in A \setminus B; \tag 6$

if

$b = f(a) \in B, \tag 7$

then

$f(b) = f(f(a)) = f(a) = b; \tag 8$

we note

$B \ni b \ne a \in A \setminus B, \tag 9$

which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.