For a function $f:X\rightarrow Y$ we have $f(U)\subset V\iff U\subset f^{-1}(V)$ proof For a function $f:X\rightarrow Y$ we have $f(U)\subset V\iff U\subset f^{-1}(V)$ proof
I'm following a book and it just uses this, it doesn't say anything about the function, so I've not assumed it's one-to-one. I've spent to much time on this (as in it should be 4 lines for both ways at most) so I hope you can help me.
What have I tried?
For $\implies$
$x\in U\implies f(x)\in f(U)$ then by hypothesis $f(x)\in V$
Here's where I get stuck, because $f^{-1}(\{f(x)\})\subset f^{-1}(V)$ isn't true.
Wait, It is because $f(x)\in V$ everything that maps to $f(x)$ will be in $f^{-1}(V)$, writing this really does help!
I then say $x\in f^{-1}(\{f(x)\})$ thus $x\in f^{-1}(V)$ - is that right?
For $\impliedby$
$y\in f(U)\implies \exists x\in U:f(x)=y$ thus $x\in f^{-1}(V)$ by hypothesis, so $f(x)\in f(f^{-1}(V))$ which is the same as $y\in V$
I don't like these sort of proves largely because they're glossed over in the first chapter of all books, and the term "function" is used very ambiguously (is it one-to-one implicitly for example?)
Are these proofs okay?
 A: $$
\Rightarrow: U\subset f^{-1}(f(U))\subset f^{-1}(V) 
$$
$$
\Leftarrow: f(U)\subset f(f^{-1}(V))\subset V
$$
Just read through (maybe more than once) the first chapter of James Munkres, Topology. Its the best to me. 
A: First of all note that by definition of the preimage of $V\subset Y$ under $f:X\rightarrow Y$ we have: $$x\in f^{-1}\left(V\right)\iff f\left(x\right)\in V$$

$\Rightarrow$
$x\in U\Rightarrow f\left(x\right)\in f\left(U\right)\subset V\Rightarrow x\in f^{-1}\left(V\right)$. Application of definition
$\Leftarrow$
$y\in f\left(U\right)\iff\exists x\in U\; f\left(x\right)=y$.
Here $x\in U\subset f^{-1}\left(V\right)$ so that $y=f\left(x\right)\in V$ again applying definition.
A: Well the proofs seem okay. But it can be considerably shortened. I'm sure somewhere in the book you use, you will find the following definitions, 
$$ f^{-1}(A) = \{ x \ | \ f(x) \in A  \} $$
$$ f(B) = \{f(x) \ | \ x \in B\} $$
$\implies$
$x \in U \implies f(x) \in f(U) \implies f(x) \in V$ from which we can directly concluse according to the definition above that $x \in f^{-1}(V)$ 
$ \Leftarrow$
Lets use a trick. Suppose $f(x) \in f(U)$ which is entirely fair again by the definition. Then we have $x \in U \implies x \in f^{-1}(V) \implies f(x) \in V$
And we are done. The term "function" is rarely used ambiguously. The function $f$ need not be one-to-one for this result to work. I believe you were thrown off by the notation $f^{-1}(V)$ but its definition does not stipulate the existence of an inverse function. 
