# If $f\in\hbox{Hom}_{\mathbb{Z}}(\prod_{i=1}^{\infty }\mathbb{Z},\mathbb{Z})$ and $f\mid_{\bigoplus_{i=1}^{\infty } \mathbb{Z}}=0$ then $f=0$.

Prove that if $f\in \hbox{Hom}_{\mathbb{Z}}(\prod_{i=1}^{\infty }\mathbb{Z},\mathbb{Z})$ and $f\mid_{\bigoplus_{i=1}^{\infty } \mathbb{Z}}=0$ then $f=0$.

I took an element of $\prod_{i=1}^{\infty }\mathbb{Z}$, that is, $(m_{1},m_{2},…)$, and because $f$ is homomorphism we have $f(m_{1},m_{2},…)=f(m_{1},0,0,…)+f(0,m_{2},0,…)+\cdots$ and because $f$ is zero on $\bigoplus_{i=1}^{\infty } \mathbb{Z}$ so $f(m_{1},m_{2},…)=0$, but I am not sure that my solution is right. Please tell me if it is right or wrong, and if it is wrong, please help me to make it right. Thank you.

#### Solutions Collecting From Web of "If $f\in\hbox{Hom}_{\mathbb{Z}}(\prod_{i=1}^{\infty }\mathbb{Z},\mathbb{Z})$ and $f\mid_{\bigoplus_{i=1}^{\infty } \mathbb{Z}}=0$ then $f=0$."

Let $f:\Bbb Z^{\Bbb N}\to\Bbb Z$ such that $f$ is zero on $\Bbb Z^{(\Bbb N)}$ (the direct sum of countable copies of $\Bbb Z$), and let $x=(x_n)_{n\ge 0}\in\Bbb Z^{\Bbb N}$. We want to prove that $f(x)=0$. Write $x_n=2^nu_n+3^nv_n$ with $u_n,v_n\in\Bbb Z$. (We can do this since $\gcd(2^n,3^n)=1$ for all $n\ge 0$.) Set $u=(2^nu_n)_{n\ge 0}$ and $v=(2^nv_n)_{n\ge 0}$. Then $u,v\in \Bbb Z^{\Bbb N}$ and $f(x)=f(u)+f(v)$. But $$f(u)=f(u_0,2u_1,\dots,2^{n-1}u_{n-1},0,\dots)+f(0,\dots,0,2^nu_n,\dots).$$ Since $f$ is zero on $\Bbb Z^{(\Bbb N)}$ we get $f(u_0,2u_1,\dots,2^{n-1}u_{n-1},0,\dots)=0$, so $f(u)=f(0,\dots,0,2^nu_n,\dots)$. But $f(0,\dots,0,2^nu_n,\dots)=2^nf(0,\dots,0,u_n,2u_{n+1},\dots)$ and therefore $2^n\mid f(u)$ for all $n\ge 0$, so $f(u)=0$. Analogously $f(v)=0$.