Determining the space complexity of $x$-obstruction-free $k$-set agreement for $x\leq k$ is an open problem. In $x$-obstruction-free protocols, processes are required to return in executions where at most $x$ processes take steps. The best known upper bound on the number of registers needed to solve this problem among $n>k$ processes is $n-k+x$ registers. No general lower bound better than $2$ was known. We prove that any $x$-obstruction-free protocol solving $k$-set agreement among $n>k$ processes uses at least $\lfloor(n-x)/(k+1-x)\rfloor+1$ registers. Our main tool is a simulation that serves as a reduction from the impossibility of deterministic wait-free $k$-set agreement: if a protocol uses fewer registers, then it is possible for $k+1$ processes to simulate the protocol and deterministically solve $k$-set agreement in a wait-free manner, which is impossible. A critical component of the simulation is the ability of simulating processes to revise the past of simulated processes. We introduce a new augmented snapshot object, which facilitates this. We also prove that any space lower bound on the number of registers used by obstruction-free protocols applies to protocols that satisfy nondeterministic solo termination. Hence, our lower bound of $\lfloor(n-1)/k\rfloor+1$ for the obstruction-free ($x=1$) case also holds for randomized wait-free free protocols. In particular, this gives a tight lower bound of exactly $n$ registers for solving obstruction-free and randomized wait-free consensus. Finally, our new techniques can be applied to get a space lower of $\lfloor n/2\rfloor+1$ for $ε$-approximate agreement, for sufficiently small $ε$. It requires participating processes to return values within $ε$ of each other. The best known upper bounds are $\lceil\log(1/ε)\rceil$ and $n$, while no general lower bounds were known.