Munkre's Topology Ch 1 sec. 2, ex. #1:

[benorin]

Summary:: Subset of Codomain is Superset of Image of Preimage, and similar proof for subset of domain

I was having a hard time doing the intro chapter's exercises in Munkres' Topology text when last I worked on it, and I just wanted to make sure that there's nothing betwixt analysis and topology I'm missing? It's been a while since college for me, so perhaps I'm just a bit forgetful of things disused for two decades. The exercise I got stuck on was

Munkres' Topology Ch 1 sec. 2, ex. #1:

If $f: A\rightarrow B$ and $A_0\subset A$ and $B_0\subset B$. (a) show that $A_0\subset f^{-1}(f(A_0))$ and that equality holds if $f$ is injective. (b) show that $f(f^{-1}(B_0))\subset B_0$ and that equality holds if $f$ is surjective.

I even found a proof (proofwiki.org) that is essentially the same problem. Couldn't follow the flow of the proof much at all; there didn't seem to be a rhyme nor reason to the starting and ending points of the proof, why one seemly identical definition differing only in notation to the next line was considered and step worth writing out. I'm typically able to follow even protracted $\epsilon , \delta -$proofs like in real analysis but this simple, basic algebra type proof is giving me trouble I did not expect. Munkres did mention that students would feel comfortable in the beginning of the chapter and find their expertise evaporating near the middle of it.

 

[member 587159]

Let us start with proving $A_0\subseteq f^{-1}(f(A_0))$.

You prove these things by showing that $x\in A_0\implies x\in f^{-1}(f(A_0))$.

So, let us fix $x\in A_0$. You have to show that $x\in f^{-1}(f(A_0))$. By definition, this means that $f(x)\in f(A_0)$. But since $x\in A_0$, trivially $f(x)\in f(A_0)$ so this is really writing out what the definitions mean.

More direct, if $x\in A_0$, then $f(x)\in f(A_0)$, hence $x\in f^{-1}(f(A_0))$.

Another way to think about this, which I use, is just try to understand in words what is happening: $f^{-1}(f(A_0))$ is the set of all points in the domain that get mapped by $f$ to $f(A_0)$. Clearly $A_0$ is contained in this set.

Were you able to understand this? If yes, can you try to prove $f(f^{-1}(B_0))\subseteq B_0$ on your own now?