So I devised this proof that 1=0. Of course it is false, but I don't know why. Why?
x+1=y
(x+1)/y=1
((x+1)/y)-1=0
((x+1)/y)-(y/y)=0
x-y+1/y=0
x-y+1=0
x-y+1=x-y+1/y
y(x-y+1)=x-y+1
y=1
x+1=1
x=0 **
y-1=x
(y-1)/x=1
((y-1)/x)-1=0
((y-1)/x)-(x/x)=0
y-x-1/x=0
y-x-1=0
y-x-1=y-x-1/x
x(y-x-1)=y-x-1
x=1 *...
@Gigili by accident I've pressed "review" button instead of "users". Wanted to show you my second account user. He has totally different gravatar. But you never know if I lie or not.
So a functor preserving $0 \to M \to \dots$ as $0 \to Hom(M,N) \to \dots$ is left exact (and possibly also exact). What sort of exact is a functor that turns $0 \to M \to \dots$ into $\dots \to Hom(M,N) \to 0$?
@MattN I don't understand the question... Left exact means that kernels are preserved, right exact that cokernels are preserved. But what do you mean with the "What sort of exact" part?
Left exactness of $F$ is the following statement: whenever $k: K \to A$ is a kernel of $f: A \to B$ then $F(k): F(K) \to F(A)$ is a kernel of $F(f): F(A) \to F(B)$.
@tb Does it follow from the properties of a functor that it has to be contravariant or can I have a functor $F$ from $R$-modules to itself such that $F: M \mapsto Hom(M,N)$ and $F: (f: M \to M^\prime) \mapsto (F(f): Hom(M,N) \to Hom(M^\prime, N))$?
@Gigili 7 bronze 3 silver and 3 gold, check his home page, that is the only question he has asked, and a reasonable assumption that he hadn't had deleted questions earlier.
How many digits of $\pi$ are currently known?
