Is pseudorandom function also a one-way function? Can I assume this? Specifically, I want to know if the following two cases are valid. Suppose $prf_k(m)=c$


One-wayness: Only given $c$, we cannot reveal $m$.
Without key $k$, even given $m$, cannot get $c$.


They seem intuitive, but I want to be assured of their validity when I say that. Is a pseudorandom function also a one-way function?
(Note the OP's clarification that even for 1, the adversary does not have the key k.)

Yes. $\:$ The way to show anything like those, is to use the fact that oracle access to prfk

is sufficient to efficiently determine the extent to which an adversary is violating those.

def straight-line_reduction_1_():
 m = random_element_of_(domain(F))
 query F on m
 let c be the output of F on m
 send c to inner_adversary_1
 forward any queries and responses between inner_adversary_1 and the F oracle
 let guess be the output of inner_adversary_1 on c
 if guess == m:
  return True
 else:
  return False


straight-line_reduction_1_ uses only a small amount of space and makes only one more oracle query than inner_adversary_1, and straight-line_reduction_1_'s runtime is dominated by forwarding queries/responses from/to inner_adversary_1. ​ (So, straight-line_reduction_1_

is efficient.) ​ For all F, the probability of straight-line_reduction_1_ outputting True is equal

to the probability of inner_adversary_1 correctly guessing m given c. ​ However, if F is a

truly random function and inner_adversary_1 makes at most q queries, then the probability

of inner_adversary_1 correctly guessing m given c is at most (q+1)/(cardinality(domain(F))).

(That should be proven, but I'll only bother doing so if you ask.)

Given that result (which I didn't provide a proof of), it then follows that the probability of inner_adversary_1 correctly guessing m given c by making at most q queries is at most (q+1)/(cardinality(domain(F))) plus the extent to which the combination of inner_adversary_1

with straight-line_reduction_1_ distinguishes the function family from random.

In particular, if the function family is pseudorandom then 1 holds.

One can proceed analogously to show that

if the function family is pseudorandom then 2 also holds.