If L = {ww : w ∈ L(1*01*)} it means that w = $1^a$0$1^b$ and ww = $1^a$0$1^b$$1^a$0$1^b$
If I want to prove that this language is context-free by giving a context-free grammar, can I give a CF grammar for $1^a$0$1^a$$1^b$0$1^b$? Would it be the same thing? I'm switching the order of the inner 1's, since if a = 2 and b = 3, there will be 5 inner 1's and it doesn't seem to matter if its the 2 a's or the 3 a's that come first.
As long as you replace $1^b1^a$ with $1^a1^b$ or $1^{a+b}$, and not a new free parameter like $1^c$, then yes it is the same language. So all you need to do is to be able to generate $1^a01^a$ for any $a$, and then by starting with the rule $S \to AB$ where $A$ and $B$ have the same grammar representation that gives $1^a01^a$ or $1^b01^b$ for arbitrary $a,b$ then you are done.