Is there a surjective, non-injective group homomorphism (under addition) from $\mathbb{C}$ to $\mathbb{C}$?

Question

Is there a surjective, non-injective group homomorphism (under addition) from $\mathbb{C}$ to $\mathbb{C}$?

It need not be continuous.

Answer 1

It is well-known that $\Bbb C$ is a vector space over $\Bbb Q$. Let $B$ be the basis of this vector space. Now, let $A$ be a countable subset of this basis. From this, we can label the elements of $A$ as a sequence $a_n$ for any $n \in \Bbb N$. The basic trick behind this problem is that you can shift the whole sequence over by one term so that $a_{n+1}$ gets sent to $a_n$ and that $a_1$ gets sent to 0, which creates a surjective but not injective endomorphism.

Using these definitions, we can now define an endomorphism $g : \Bbb C \rightarrow \Bbb C$ by saying that $g$ leaves $\Bbb Q$ fixed and defining how $g$ acts on $B$. First, for any $x \in B$ such that $x \notin A$, $g(x)=x$ (i.e. $g$ leaves $B-A$ fixed). Then, for any $a_n \in A$ some integer $n > 1$, then $g(a_n)=a_{n-1}$. However, $g(a_1)=0$.

Now, clearly $g(0)=0$ (since $g$ leaves $\Bbb Q$ fixed) and $g(a_1)=0$, so $g$ is not injective. However, $g$ is surjective because for any $x \in B$, if $x \notin A$, $g(x)=x$ and if $x \in A$, then $x=a_n$ for some integer n and $g(a_{n+1})=x$. Thus, $g$ contains all of $B$ in its range and all of $\Bbb Q$ in its range (since $g$ leaves $\Bbb Q$ fixed), so since $g$ is a homomorphism, $g$ has all of $\Bbb C$ in its range. Therefore, $g$ is a surjective but not injective endomorphism on the additive group of $\Bbb C$.