Well, I've always thought and always been brought up on the fact that the word "piss" is in fact swearing.
I took it out of this question, then asked the OP not to swear. He came back and insisted it wasn't swearing, after googling it seems this differs in different parts of the world.
What is ...
@Pissedofflayman I'd advise you change your username. I'd also advise you stop spamming chatrooms with the same video. Take a look at chat.stackexchange.com/faq while you're at it too.
@KajHansen this thing is for sets. So, it shouldn't be that weird.
@KajHansen $f : A \rightarrow B$ is epimorphism iff for all sets $Z$ and for all $\alpha_1,\alpha_2 : B \rightarrow Z$ such that $\alpha_1 \circ f = \alpha_2 \circ f$ implies $\alpha_1 = \alpha_2$
One side is easy if f is surjective then it has right inverse so it is an epimorphism.
Suppose $f : A \rightarrow B$ is a epimorphism. Suppose f isn't surjective. That is there exists $b \in B$ such that there exists no $a \in A$ that gets mapped to it. Pick arbitrarily fixed function $\alpha_1 : B \rightarrow A$ construct $\alpha_2 B \rightarrow A$ as follows $\alpha_2$ is same as $\alpha_1$ however $\alpha_1(b) \neq \alpha_2(b)$. Then $\alpha_1 \circ f = \alpha_2 \circ f$ but $\alpha_1 \neq \alpha_2$
that is what I did @arctictern
I want a direct approach though
I am not fully content with this prove by contradiction.
The above contradicts the fact that $f$ is an epimorphism so f must be surjective.
I thought maybe I can construct somehow $\alpha_1$ and $\alpha_2$ from $B \rightarrow f(A)$ somehow and prove that $f(A) = B$ somehow using fibers or something like that
@Pissedofflayman I better leave just to be sure. I don't want to get in worse trouble. I've been getting in trouble around here a lot lately. They said one more mess up and I'd be banned...
I wish I could remember the one property I'm thinking of. I've even seen an algebraic proof of the fact, but can not wrap my head around wtf is going on.
@DHMO suppose his set contains an element, since it is open it has to contain a neighbourhood of this point, then it has to contain the limit points of this neighbourhood, etc... Then it has to contain the whole of R
Without loss of generality, I can assume it depends on $x$. (If it only depends on $y$, then I can swap the names of the variables; if it depends on neither, it's a constant map and therefore has closed image.)
If f us unbounded above and below hen the image is closed. Suppose f>=M for all x,y then we're trying to find a polynomial such that there's no f(x,y)=M but f gets arbitrarily close but that isn't possible
Problem with what I've just said is that the image could degenerate to a point.
I'm fine with you announcing it. If it turns out to be a valid counterexample I'm going to be shocked; if it turns out to hinge on a technicality, I'm going to be irritated.
@Sophie I agree with you that as long as you go to infinity near infinity the image is closed. But it's not clear that this is true for multivariable polynomials.
@DHMO I think this is a little less inspiring when it's something you got off MO or whatever as opposed to having independent personal interest.
One could probably make a fairly interesting talk on FLT that just focuses on how hard it is to prove various cases, e.g. what's the smallest $n$ for which there's no elementary proof.
The post might be sincere crackpottery. OP kinda appears to have some mission, as if it to bait contention to later throw out their non-mainstream point.