Question 1: $$0 \to A\mathop \to \limits^f B\mathop \to \limits^g C \to 0$$ is an exact short sequence,in order to prove $$\cdots \to H^q (A)\mathop \to \limits^{f^* } H^q (B)\mathop \to \limits^{g^* } H^q (C)\mathop \to \limits^{d^* } H^{q + 1} (A) \to \cdots$$ is an exact long cohomology sequence,we need to prove$${\mathop{\rm Im}\nolimits} f^* = \ker g^* $$,I can prove $$ {\mathop{\rm Im}\nolimits} f^* \subset \ker g^* $$ because $${\mathop{\rm Im}\nolimits} f = \ker g$$,but how to prove $ {\mathop{\rm Im}\nolimits} f^* \supset \ker g^* $? \\ \\ \\ Question 2: What makes the cohomology groups different by digging 2 points of $R^2$ from that of $R^2$ ?How does the closed and exact differential forms change? Thank you! the visualized questions are also in the attach files
A1: Simple diagram chasing. Take in H^{q}(B), b in Ker d such that g*=[g(b)]=0. I.e. g(b)=dc for some c in C^{q-1}(C). To show: there exists a in C^{q}(A) and b' in C^{q-1}(B) such that b-db'=f(a). Choose b' in g^{-1}(c) (which exists since g is surjective). Note that since g is a chain map, g(b-db')=g(b)-dg(b')=dc-dc=0. So, since Ker(g)=Im(f), there exists a in C^{q}(A) such that f(a)=b-db'. A2: This question is kinda vague, but notice that by digging points in R^2, new forms appear. Here is an example that illustrated the nature of the phenomenon: Consider on R²-0 the 1-form (-ydx+xdy)/(x²+y²) [<---ill defined on all of R²]. This form is closed but it is not exact. This is because it is equal to [itex]d\theta[/itex] everywhere where [itex]\theta[/itex] (the polar angle) is defined (i.e. R² - {a half-ray}). Were (-ydx+xdy)/(x²+y²) exact, this would imply that there exists a differentiable extension of the polar angle function [itex]\theta(x,y)[/itex] to all of R²-0. But clearly, there exists no such even continuous extension.