My discovery about mathematics

                                1

The definition of the mapping by the graph has defect.
The theory of ⊗ isn't consistent if you define the mapping by the graph.

                                2

Let M be a module. Define f: M∋x→x+x∈M. Put F=Z/2Z and i=1_F.
Let G(g)={(x,g(x)) |(x∈A)∧(g(x)∈B)} be the graph of the mapping g: A∋x→g(x)∈B.
G(g⊗i)={(x⊗(1+2Z),y⊗(1+2Z)|(x,y)∈G(g)}.
Theorem 1. If you define the mapping by the graph, h: M∋x→x+x∈Im(f) ⇒ h=f.
Proof. G(f)={(x,x+x) |(x∈M)∧(x+x∈M)}={(x,x+x) | x∈M}={(x,x+x) |(x∈M)∧(x+x∈Im(f))}
=G(h). So h=f.♦
By the axiom of the equality, f=h ⇒ Im(f⊗i)=Im(h⊗i). (You will know that this is wrong.)
Theorem 2. If you define the mapping by the graph,the theory of ⊗ isn't consistent.
Proof. Put M=Z. M⊗F≅F. You may define M⊗F=F and x⊗(y+2Z)=xy+2Z in M⊗F=F.
f⊗i: M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈M⊗F. f⊗i: F∋x+2Z→2x+2Z=0+2Z∈F. Im(f⊗i)≅{0}.
Im(f)⊗F=(2Z)⊗F≅F. You may define Im(f)⊗F=F and 2x⊗(y+2Z)=xy+2Z in Im(f)⊗F=F.
By theorem 1, f: M∋x→x+x∈Im(f). f⊗i:M⊗F∋x⊗(1+2Z)→(x+x)⊗(1+2Z)∈Im(f)⊗F.
f⊗i: F∋x+2Z→2x⊗(1+2Z)=x+2Z∈Im(f)⊗F=(2Z)⊗F=F. Im(f⊗i)=F. F≅{0}.
This is contradiction.♦
Theorem 3. The set theory which assumes the axioms of ZF and existence of the module Z
and defines the mapping by the graph isn't consistent.
Proof. Put M=Z. M, Im(f)=2Z={x+x |x∈M}, 1+2Z={x|(x∈M)∧(x∉2Z)}, F={2Z, 1+2Z}
and G(f)=G(h)={(x, x+x) |x∈M} are sets. So f=h.
The contradiction in the proof of theorem 2 is led when you define M⊗F=F and
x⊗(1+2Z)=x+2Z in M⊗F=F and (2Z)⊗F=F and (2x)⊗(1+2Z)=x+2Z in (2Z)⊗F=F. ♦
Theorem 4. The theory treating the free modules isn't consistent if you define the mapping
by the graph.
Proof. Z is a free module. M⊗N=Z^(M×N)/(0_M⊗0_N). Z^(M×N) and 0_M⊗0_N are free modules.♦
You may not be able to get the tensor products and the projective resolutions of the
modules in the consistent theory.

                                   3

You may have to reexamine the theorems proved by using ⊗.
Many conjectures may be back to the starting points or may become nonsense.