Let $p$ be prime and the $\mathbb{Z}$-module $M=\prod_{n=1}^\infty\mathbb{Z}_{p^n}$ where $\mathbb{Z}_{p^n}$ is $\mathbb{Z}$ modulo $p^n$. Define a map $f : M\to M$ by $f(a_1 + p\mathbb{Z},a_2 + p^2 \mathbb{Z},a_3 + p^3 \mathbb{Z},\ldots) = (0,pa_1 + p^2 \mathbb{Z},pa_2 + p^3 \mathbb{Z},\ldots)$
The question: Is the map $f$
(1) surjective,
(2) injective?
Answer: The image of $f, f(M)=Mp$. This implies that $f$ is a multiplication map by $p$ and so is injective. However I am failing to verify whether it is surjective or not.
The map $f$ is not surjective, because any element in the image of $f$ has zero first component.
Checking injectivity requires a little bit of work. $f$ is a group homomorphism, so let's check that the kernel of $f$ is zero. Suppose $f(a_1 + p\mathbb{Z},a_2 + p^2 \mathbb{Z},a_3 + p^3 \mathbb{Z},\ldots) = (0,pa_1 + p^2 \mathbb{Z},pa_2 + p^3 \mathbb{Z},\ldots)$ is the zero vector. So $pa_1\equiv 0\pmod{p^2}$. Dividing by $p$ we see that $a_1$ is a multiple of $p$. Similarly any $pa_i\equiv 0\pmod{p^{i+1}}$ and $a_i$ is a multiple of $p^i$. So the initial vector $(a_1, a_2,...)$ is the zero element of $M$.