Proving Functions: The Relationship Between Onto and 1-to-1 Properties

[SqrachMasda]

everything i have to prove seems impossible
then i see it done and it seems so easy
any help on starting a proof,...anybody ever have this problem
i am new to this stuff, but my problem is I don't know what i need to show and what is legal to use
so this is my current problem, I'm sure it's extremely easy for all you math geniuses so spare me the arrogance

If f itself is 1-to-1 then f^-1 is a function.
Let f:G->H be an onto function
prove
(i) If V is contained in H then f(f^-1(V))=V

i'm pretty sure i understand onto and 1-1 definition
i also understand why it's true, i think
i also can somewhat understand the homomorphism stuff of it

i just don't know where to start, what to do, what it really is asking

then i got to prove this, but have not looked into it because I'm still on (i)
(ii) If W is contained in G then W is contained in f^-1(f(W)) and this could be a strict containment

abstract algebra seems to be an endless barrier in my math knowledge
is the key to it, memorizing endless amounts of definitions??
seriously

 

[rachmaninoff]

1) Write down the definitions. (that's how you start!)

 

[SqrachMasda]

i did, i wrote evrerything down i knew
but I'm not sure how to apply it i guess, (that's how i started!)

 

[quasar987]

anatomy of a proof

As you see more and more proofs, you'll start to see that there is a finite number of tehcniques one uses to prove things.

But almost all of them begin by writing down the definitions.

Let f:G->H be an onto function
prove
(i) If V is contained in H then f(f^-1(V))=V

Start with the definition of an onto function from G to H.

Follow with the definition of f^-1(V).

Then with the definition of f(A) where A in a subset of G.

Then replace A by f^-1(V).

Just by applying the definitions you'll have what f(f^-1(V)) is. And if it is indeed V, that's what you'll find.

 

[SqrachMasda]

thanks, i think that may make sense breaking it all down like that

 

[mathwonk]

heres one suggestion: try to refrain from being defensive and rude to people you are asking for help. i was going to help but was put off by the "spare me the arrogance" comment.

 

[JasonRox]

heres one suggestion: try to refrain from being defensive and rude to people you are asking for help. i was going to help but was put off by the "spare me the arrogance" comment.

Wow, I didn't even notice that until you mentionned it.

Those are harsh words.

 

[matt grime]

...i'm sure it's extremely easy for all you math geniuses so spare me the arrogance...

If f itself is 1-to-1 then f^-1 is a function.

no, that is not true. only if f is a bijection (one to one and onto) is there an inverse function. Perhaps that sentence ought to have come after this one.

Let f:G->H be an onto function

prove
(i) If V is contained in H then f(f^-1(V))=V

the most useful way to prove set equality is to show each is contained in the other. Pick something in V, and shwo that it is in f(f^-1(V)), which is obvious since for v in V v=ff^-1(v) by the definition of inverse function: g is inverse to f if and only if fg and gf are the identity maps. The other containment is as easy to show. Arguably too easy in some sense and it genuinely isn't easy to write the proof in these cases: it isn't clear what constitutes a solid argument and what doesn't when the solid argument is very close to just saying 'well, it is'.

i also can somewhat understand the homomorphism stuff of it

where do homomorphisms come into anything here?

(ii) If W is contained in G then W is contained in f^-1(f(W)) and this could be a strict containment

if you understand the inverse image of a set you will see that obviously more than just W may map into f(W) (we're now droppping the hypotheses on f being injective I take it) and certainly W maps into f(W). Provide a counter example for the second part.

abstract algebra seems to be an endless barrier in my math knowledge
is the key to it, memorizing endless amounts of definitions??
seriously

you've memorized thousands of words; i suggest you didn't hat learning french at school because it required vocabulary. why should this be any different?
as it is you only need to memorize very few things compared to the knowledge you'll be required to keep in your head for everything else in this life.