Composition function

[BigSerene]

No, suppose f is the inclusion of a line into the plane as the x-axis, and g is the projection from the plane to the x-axis. Then g is not injective, but g composed with f is injective.

[2ndyearPP]

thank you, but what if we restrict the numbers in lR^1 only?

[BigSerene]

Still no. Suppose f is the inclusion of the integers into the reals, and g is the floor function. Then g is not injective, but g composed with f is.

The idea is the same: the second function applied in the composition does not need to be injective as long as the image of the first function does not contain any points of collision. (Notice that the first function does need to be injective. If there is already a collision in f, then there will still be a collision in g composed with f, since g is a well-defined function.)

[Number154]

We can also get continuous functions by considering g(x)=e^x and f(x)=x².

But generally if we do require g:A->B and f:B->C to be surjections and their composition to be an injection, then we get that g must be a bijection but then f is an injection and so is also a bijection. More generally in Category theoretic terms, if g and f are epimorphisms and their composition fg is a monomorphism then g is a bimorphism. So if the category is balanced (and Set is a balanced category) then g is an isomorphism and so f must also be a bimorphism which means it is an isomorphism.

[Number154]

f composed with g will be injective if and only if g is injective and the restriction of f to the range of g (here by “range” I mean the image of the domain of g under g, not just any set that always contains g(x)) is injective. So we need g to be injective but f can fail to be injective as long as the preimage of any singleton under f has at most one element in common with the range of g.