There is no injective function from X to the set of natural numbers.

It also follows from this construction that i is injective.

If the function is injective the original can be reconstructed.

In fact no function of any kind from A to B is injective.

The map f itself need not be injective, only its derivative.

It is not difficult to show that f is injective.

There are multiple other methods of proving that a function is injective.

A function f with a left inverse is necessarily injective.

So in particular, every abelian group is subgroup of an injective one.

Formally, this follows from the fact that the code is an injective map.

T λ is injective but does not have dense range.

In mathematical terms, it is a total injective function.

Even if the mapping is injective its inverse will not be continuous.

For a separated presheaf, the first arrow need only be injective.

This action is faithful, which is equivalent to being injective.

The representation is said to be faithful if it is injective.

In concrete categories, a function that has a left inverse is injective.

A continuous injective map can't create a boundary, giving us our contradiction.

More generally, every injective module is algebraically compact, for the same reason.

Two conditions will be imposed on (to be injective, and generating).

We require f to be injective in order for the inverse function to exist.

However, every image of a perfect space under an injective continuous map is perfect.

Assume for a contradiction that h is not injective.

Every object in a Grothendieck category has an injective hull.

An operator may be injective, even bounded below, but not invertible.