Product of sets of semi-infinite Configurations "equals" set of bi-infinite configurations? Assume $\mathcal{A}$ is a finite set ("Alphabet") equipped with the discrete topology. Then let $\mathcal{A}^{\mathbb{Z}}$ be the product space of bi-infinite configurations and equip it with the associated product topology.
In the same sense, let $\mathcal{A}^{\mathbb{Z}_{\leq 0}}$ be the set of the left-infinite configurations and $\mathcal{A}^{\mathbb{Z}_{>0}}$ the set of right-infinite configurations equipped with the associated product topology, respectively.
(1) Do we then have that
$$
\mathcal{A}^{\mathbb{Z}_{\leq 0}}\times\mathcal{A}^{\mathbb{Z}_{>0}}\simeq\mathcal{A}^{\mathbb{Z}}?
$$
(2) A somehow related question I have is whether
$$
\left(\mathcal{A}^{\mathbb{Z}_{\leq 0}}\times\mathcal{A}^{\mathbb{Z}_{>0}}\times\mathbb{Z}\right)\cup\left(\mathcal{A}^{\mathbb{Z}}\times\left\{-\infty,+\infty\right\}\right)\simeq \mathcal{A}^{\mathbb{Z}}\times\bar{\mathbb{Z}},~~\text{ where  }\bar{\mathbb{Z}}=\mathbb{Z}\cup\left\{\pm\infty\right\}.
$$

My naive answer to (1) would be 'yes' since I guess that in fact, we have equality for the Cartesian products, i.e.
$$
\mathcal{A}^{\mathbb{Z}_{\leq 0}}\times\mathcal{A}^{\mathbb{Z}_{>0}}=\mathcal{A}^{\mathbb{Z}}.
$$
Hence, if we equip both sides with the same topology, the identity map should give a topological isomorphism.
 A: As you noticed, you have equality of sets, and so, in order to prove the homeomorphism, it is enough to show that you have the same open sets, or equivalently, the same basic open sets.
In the product topology you can take as a basis of $X^I$, where $X$ is any topological space and $I$ any set, the family of sets
$U = \prod_{i \in I} U_i$,
where each $U_i$ is an open set of $X$ and $\{ i \in I : U_i \neq X \}$ is finite.
Hence, if $U$ is a basic open set of $\mathcal{A}^{\mathbb{Z}}$, then $U = \prod_{i \in \mathbb{Z}} U_i$, and $\{ i \in \mathbb{Z} : U_i \neq \mathcal{A} \}$ is finite.
You can re-write $U$ as 
$\prod_{i \in \mathbb{Z}_{\leq0}} U_i \times \prod_{i \in \mathbb{Z}_{>0}} U_i$,
and you certainly have that
$$\{ i \in \mathbb{Z}_{\leq0} : U_i \neq \mathcal{A} \} \text{ and } \{ i \in \mathbb{Z}_{>0} : U_i \neq \mathcal{A} \}$$
are both finite, yielding that $\prod_{i \in \mathbb{Z}_{\leq0}} U_i$ is an open set of $\mathcal{A}^{\mathbb{Z}_{\leq0}}$
and $\prod_{i \in \mathbb{Z}_{>0}} U_i$ is an open set of $\mathcal{A}^{\mathbb{Z}_{>0}}$.
It follows that $U$ is an open set of $\mathcal{A}^{\mathbb{Z}_{\leq0}} \times \mathcal{A}^{\mathbb{Z}_{>0}}$.
For the converse, use the fact that the union of finite sets is finite.
