>>11525985
Naturally.
By the way, would you happen to have a good proof that the homology is non-trivial?
Mine sorta goes like this:
>[math]V[/math] is open, so there's an open ball [math]B[/math] around [math]f(p)[/math] whose closure is contained entirely in V
>we take the sphere [math]\partial B[/math]
> [math] \partial B[/math] has a non-trivial homology class in [math]\mathbb{R}^n - \{ f(p) \} [/math]
>Poincaré duality followed by the de Rham isomorphism gives us a differential form [math]\omega [/math] defined in [math]\mathbb{R}^n - \{ f(p) \} [/math] with [math]\int_{ \partial B} \eta \neq 0[/math]
>we restrict [math]\omega[/math] to [math]V - \{ f(p) \} [/math] to obtain [math]\eta[/math]
> [math]\int_{ \partial B} \eta \neq 0[/math], trivially, which completes the proof that homology is non-trivial
Which, while extremely simple, feels unnecessarily long, but I don't really recall what was the theorem which you used for this trick.
>>11528526
> [math]g[/math] needs to be continuous
Pretty sure it doesn't, the limit [math]\lim _{x \rightarrow c} g(x)[/math] just has to exist.
>do you have a proof
I have a very shitty but sort of neat proof. Set [math]h(x) = g(x)[/math] if [math]x \neq c[/math], [math]h(c)= \lim_{x \rightarrow c}g(x)[/math]. Then, [math]h[/math] is continuous at [math]c[/math] and the proof (for [math]h[/math] ) follows the way you're thinking.
But neither of the sides of the expression depends on the value [math]g[/math] takes at [math]c[/math], so it follows.
If you insist on it, I can do an epsilon-delta one, but it's kind of a pain.
>>11529229
>all the faces have different colours
Man, that really makes it easier.
So, imagine you're holding the cube. First, you choose a face, which you hold in your direction.
Choose another face, which can't be the opposite one, and rote the cube so that it points upwards.
This means that the cube has 6*4=24 choices of positioning, so you just compute all the arrangements and divide by that.