Are all injective endomorphisms of a module automorphisms?

As far as I understand, an automorphism is an isomorphism from a set to itself. If we have a homomorphism $f:M\rightarrow M$, then, from the first isomorphism theoreom, $im(f)$ is a submodule of $M$. As it is injective, the kernel is zero, and hence the image is isomorphic to $M$.

Does this not show that $f$ is an isomorphism from $M$ to itself? What am I missing?

• Consider $\phi : \mathbb{Z} \to \mathbb{Z} : x \mapsto 2x$. – anakhro Jul 9 '15 at 16:31
• But wouldn't that have an inverse? When you apply the function, you multiply by 2. In the inverse, you divide by 2, but you only define it for even numbers. – man_in_green_shirt Jul 9 '15 at 16:33
• Or is that why you can't consider it an isomorphism? Does every element in target set need to have an inverse? – man_in_green_shirt Jul 9 '15 at 16:34
• If you define a function by "dividing by $2$", then what is the image of $1$? An isomorphism is a morphism which has an inverse; this inverse must be defined on the whole module. – Pierre-Guy Plamondon Jul 9 '15 at 16:35
• I guess the counter-intuitive fact here (but fact nonetheless) is that a module $M$ can have a strictly contained submodule $N$ which is isomorphic to $M$. The example with $\mathbb{Z}$ illustrates this. – Pierre-Guy Plamondon Jul 9 '15 at 16:44

Not in general. For example, if your base ring is $\mathbb{Z}$, and you take the map from $\mathbb{Z}$ to itself multiplying every element by $2$, this map is injective, but is not an automorphism.
However, the statement is true if $M$ is an Artinian module, as can be seen in this question.
The isomorphism theorem states that $im(f) \cong M/ ker( f )$, but this isomorphism does not need to be the injection of the submodule.
In the example above, $2\mathbb{Z} \cong \mathbb{Z}$, but not by the injection.