$\blacksquare~$ Problem:
$\bullet~$Let $A = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots~$ be the direct product of copies of $\mathbb{Z}$ indexed by the positive integers (so $A$ is a ring under component-wise addition and multiplication) and let $R$ be the ring of all group homomorphisms from $A \to A$.
$\bullet~$Let $\varphi$ be the element of $R$ defined by
$$ \varphi(a_{1}, a_{2}, a_{3}, \dots) = (a_{2}, a_{3},\dots) $$
Let $\Psi$ be the element of $R$ defined by $$ \Psi(a_{1}, a_{2}, a_{3}, \dots) = (0, a_{1}, a_{2}, a_{3}, \dots) $$
(a) Prove that $(\varphi \circ \Psi)$ is the identity of $~R~$ but $(\Psi \circ \varphi)$ is not the identity of $R$.
(b) Exhibit infinitely many right inverses for $\varphi$.
(c) Find a non-zero element $\pi$ in $R$ such that $(\varphi \circ \pi) = \mathbf{0} \in \mathbb{Z}^{\infty}$, but $(\pi \circ \varphi) \neq \mathbf{0}$.
(d) Prove that there is no non-zero element $\lambda \in R$ such that $(\lambda \circ \varphi) = \mathbf{0}.$
$\blacksquare~$ My approach: Given in problem $A = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots$
Note that, $A$ itself is a ring w.r.t component-wise addition and multiplication. And now we consider the ring $~\textbf{Hom}_{\mathbb{Z}}(A , A).~$ Also note that, $\bigg( \textbf{Hom}_{\mathbb{Z}}(A , A) , + , \circ \bigg) := (R, +, \circ)$ is our needed ring.
Let, $\varphi , \Psi $ $\in$ $\textbf{Hom}_{\mathbb{Z}}(A , A) $ such that, \begin{align*} &\varphi ( a_{1}, a_{2}, a_{3}, \dots ) = ( a_{2}, a_{3}, \dots )\\ & \Psi ( a_{1}, a_{2}, a_{3}, \dots ) = ( 0, a_{1}, a_{2}, a_{3}, \dots ) \end{align*} Where $a_{i}$ $\in$ $\mathbb{Z}$ $~\forall$ $i$ $\in$ $\mathbb{Z}^{+}$.
$\bullet~$(a) At first we need to show that, ( $\varphi \circ \Psi$ ) is the Identity map on $\textbf{Hom}_{\mathbb{Z}}(A , A)$ := $R$, but the map ( $\Psi \circ \varphi $ ) is not.
Now, according to problem we have, \begin{align*} \begin{split} (\varphi \circ & \Psi) ( a_{1}, a_{2}, a_{3}, \dots )\\ & = (\varphi( \Psi ( a_{1}, a_{2}, a_{3}, \dots )))\\ & = \varphi ( 0, a_{1}, a_{2}, a_{3}, \dots )\\ & = ( a_{1}, a_{2}, a_{3}, \dots ) \end{split} \end{align*} Therefore we have obtained that, the map $( \varphi \circ \Psi )$ is the identity mapping on $R$.
Now we'll check the case for the map $ ( \Psi \circ \varphi ) $. \begin{align*} \begin{split} (\Psi \circ & \varphi) ( a_{1}, a_{2}, a_{3}, \dots )\\ & = (\Psi( \varphi ( a_{1}, a_{2}, a_{3}, \dots )))\\ & = \Psi ( a_{2}, a_{3}, \dots )\\ & = (0, a_{2}, a_{3}, \dots )\\ & \neq ( a_{1}, a_{2}, a_{3}, \dots ) \end{split} \end{align*} Therefore, we are done!
$\bullet~$(b) We need to exhibit infinitely many inverses for $\varphi$.
Let's consider the map $\mu_{i}$ $\in$ $R$ $\forall$ $i$ $\in$ $\mathbb{N}_{0}$ , such that, \begin{align*} \mu_{i} (a_{1}, a_{2}, a_{3}, \dots) = (b_{i}, a_{1}, a_{2}, a_{3}, \dots) \end{align*} Now well observe the map $( \varphi \circ \mu_{i} )$. Let $a =$ $(a_{1}, a_{2}, a_{3}, \dots)$. \begin{align*} \begin{split} (\varphi \circ & \mu_{i} ) (a_{1}, a_{2}, a_{3}, \dots)\\ & = ( \varphi ( \mu_{i} (a_{1}, a_{2}, a_{3}, \dots) ))\\ & = \varphi ( b_{i}, a_{1}, a_{2}, \dots )\\ & = ( a_{1}, a_{2}, \dots ) \end{split} \end{align*} Therefore, we have obtained our left inverse of $\varphi$ namely $\mu_{i}$. Now as $i$ $\in$ $\mathbb{N}_{0}$, therefore, we have our set of our left inverses namely- \begin{align*} \Lambda := \{ \mu_{i} : i \in \mathbb{N}_{0} \} \end{align*} Now note that, $\rvert \Lambda \lvert $ $=$ $ \infty $. Hence we've shown that, $\varphi$ has infinitely many left inverses.
$\bullet~$(c) Let us pick a map $\pi$ $\in$ $R$, such that, for any $\textbf{a} = ( a_{1}, a_{2}, a_{3}, \dots )$. \begin{align*} \pi (\textbf{a}) = (\sigma,0 ,0 , \dots) \end{align*} Therefore, we can see that, $\pi$ is a constant map.
Now, $( \varphi \circ \pi )$ we have \begin{align*} \begin{split} ( \varphi \circ & \pi ) (\textbf{a})\\ & = ( \varphi( ( \pi(\textbf{a})) )\\ & = \varphi ( \sigma, 0, 0, \dots )\\ & = ( 0, 0, 0, \dots ) = \textbf{0} \end{split} \end{align*} But again for $( \pi \circ \varphi )$ we have \begin{align*} ( \pi \circ & \varphi ) (\textbf{a}) \\ & = ( \pi ( \varphi(a_{1}, a_{2}, a_{3}, \dots ) ) )\\ & = \pi (a_{2}, a_{3}, a_{4}, \dots)\\ & = ( \sigma, 0, 0, 0, \dots ) \neq \textbf{0} \end{align*} Hence, we're done!
$\bullet~$(d) We need to show that, $\exists$ no non-zero $\lambda$ $\in$ $R$, such that, \begin{align*} (\lambda \circ \varphi) \equiv 0 \end{align*} On the contrary let's assume there exists such a $\lambda$. And let's pick an arbitrary $\textbf{a} = ( a_{1}, a_{2}, a_{3}, \dots ) $ $\in$ $A$.
Therefore we have the following, \begin{align*} \begin{split} &(\lambda \circ \varphi) (\textbf{a}) = \textbf{0}\\ \implies & (\lambda ( \varphi (a_{1}, a_{2}, a_{3}, \dots ) ) ) = \textbf{0}\\ \implies & \lambda (a_{2}, a_{3}, a_{4}, \dots) = \textbf{0} = (0, 0, 0, 0, \dots)\\ \implies & \lambda (\hat{\textbf{a}}) \equiv \textbf{0}\\ \end{split} \end{align*} As our $\textbf{a}$ was arbitrary, we've obtained an arbitrary $\hat{\textbf{a}}$ such that, $\lambda ( \hat{\textbf{a}} ) \equiv \textbf{0}$.
Contradiction! Hence, we are done!
Can anyone please check the proof for glitches and please give new Ideas too. :)