properties of injective functions

Theorem 1.

Suppose A,B,C are sets and f:AB, g:BC are injective functions. Then the compositionMathworldPlanetmathPlanetmath gf is an injection.


Suppose that (gf)(x)=(gf)(y) for some x,yA. By definition of composition, g(f(x))=g(f(y)). Since g, is assumed injective, f(x)=f(y). Since f is also assumed injective, x=y. Therefore, (gf)(x)=(gf)(y) implies x=y, so gf is injective. ∎

Theorem 2.

Suppose f:AB is an injection, and CA. Then the restrictionPlanetmathPlanetmathPlanetmathPlanetmath f|C:CB is an injection.


Suppose (f|C)(x)=(f|C)(y) for some x,yC. By definition of restriction, f(x)=f(y). Since f is assumed injective this, in turn, implies that x=y. Thus, f|C is also injective. ∎

Theorem 3.

Suppose A,B,C are sets and that the functions f:AB and g:BC are such that gf is injective. Then f is injective.


(direct proof) Let x,yA be such that f(x)=f(y). Then g(f(x))=g(f(y)). But as gf is injective, this implies that x=y, hence f is also injective. ∎


(proof by contradictionMathworldPlanetmathPlanetmath) Suppose that f were not injective. Then there would exist x,yA such that f(x)=f(y) but xy. Composing with g, we would then have g(f(x))=g(f(y)). However, since gf is assumed injective, this would imply that x=y, which contradicts a previous statement. Hence f must be injective. ∎

Theorem 4.

Suppose f:AB is an injection. Then, for all CA, it is the case that f-1(f(C))=C.11In this equation, the symbols “f” and “f-1” as applied to sets denote the direct image and the inverse image, respectively


It follows from the definition of f-1 that Cf-1(f(C)), whether or not f happens to be injective. Hence, all that need to be shown is that f-1(f(C))C. Assume the contrary. Then there would exist xf-1(f(C)) such that xC. By defintion, xf-1(f(C)) means f(x)f(C), so there exists yA such that f(x)=f(y). Since f is injective, one would have x=y, which is impossible because y is supposed to belong to C but x is not supposed to belong to C. ∎

Theorem 5.

Suppose f:AB is an injection. Then, for all C,DA, it is the case that f(CD)=f(C)f(D).


Whether or not f is injective, one has f(CD)f(C)f(D); if x belongs to both C and D, then f(x) will clearly belong to both f(C) and f(D). Hence, all that needs to be shown is that f(C)f(D)f(CD). Let x be an element of B which belongs to both f(C) and f(D). Then, there exists yC such that f(y)=x and zD such that f(z)=x. Since f(y)=f(z) and f is injective, y=z, so yCD, hence xf(CD). ∎

Title properties of injective functions
Canonical name PropertiesOfInjectiveFunctions
Date of creation 2013-03-22 16:40:20
Last modified on 2013-03-22 16:40:20
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 22
Author rspuzio (6075)
Entry type Theorem
Classification msc 03E20
Classification msc 03E99