The professor's original question follows.
$f:A\rightarrow B$ and $|B| < |A|$. Can you make any conclusions as to whether f is injective or surjective? Can you make any conclusions as to whether $f^{-1}$ is injective or surjective?
The professor's answer follows:
f is not injective but might be surjective.
$f^{-1}$ is injective but not surjective.
My Thinking
I agree with the professor in all aspects except $f^{-1}$ is injective.
Intuitively, if f is injective then $f^{-1}$ is also injective. Plus vice versa, the converse. However, f is not injective. The converse prevents $f^{-1}$ from being injective. Where is the mistake? Am I making a mistake or him?
Note: I'm familiar with the fact that the inverse might be merely a relation and not a function.
As you said $f^{-1}$ is a symbol used for the pre-image of a function $f$. The existence of the inverse function $f^{-1}$ is linked to the concept of bijection, namely: $$f \ \text{is a bijection} \iff f \ \text{has an inverse function} \ f^{-1}$$
In this case since $f:A\to B$ domain's cardinality is strictly grater of codomain $f$ could not be injective, so $f$ could be surjective or neither (for example constant functions). So in this case $f^{-1}$ isn't a function, let's have a look at an example $$A =\{a,b,c,d\}, B =\{1,2,3\} \\ f(a)=1,f(b) = 2 , f(c)=f(d) = 3 $$
In this case $f^{-1}(3) = \{c,d\}$ and this fails the definition of a function, $f^{-1}$ could not be injective.
If we change the initial conditions $|A|< |B|$ in this case if $f$ is injective, $f^{-1}$ exist restricted to image of $f$ in $B$ and is injective. Hope this helps!