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Let $M$ be some finite set. Do we then have that $M^{\mathbb{Z}}=M^{\mathbb{Z}_{\leq 0}}\times M^{\mathbb{Z}_{>0}}$?

Since $M^{\mathbb{Z}}=\prod_{i\in\mathbb{Z}}M_i$ with $M_i=M$ for all $i\in\mathbb{Z}$, it should be possible to split this product, ie. $$ M^{\mathbb{Z}}=\prod_{i\in\mathbb{Z}}M_i=\prod_{i\in\mathbb{Z}_{\leq 0}}M_i\times\prod_{i\in\mathbb{Z}_{>0}}M_i=M^{\mathbb{Z}_{\leq 0}}\times M^{\mathbb{Z}_{>0}}. $$

Or am I completely wrong and think too naively?

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    $\begingroup$ This is correct. $\endgroup$ – Crostul Mar 31 '17 at 7:40
  • $\begingroup$ Then it should also be true that $(M^{Z_{\leq 0}}\times M^{Z_{>0}}\times Z)\cup (M^Z\times\left\{-\infty,+\infty\right\})=M^Z\times Z'$ with $Z'=Z\cup\left\{-\infty,+\infty\right\}$? $\endgroup$ – John_Doe Mar 31 '17 at 7:43
  • $\begingroup$ How exactly do you define $\prod_{i\in\mathbb{Z}}M_i$ and $M^{\mathbb Z}$? $\endgroup$ – drhab Mar 31 '17 at 7:46
  • $\begingroup$ @Crostul I would rather say that it is almost correct. $\endgroup$ – drhab Mar 31 '17 at 7:50
  • $\begingroup$ @drhab The definition I had in mind was is the following. For each $i\in I$, let $A_i$ be a set. Then $\prod_{i\in I}A_i:=\left\{a\colon I\to\cup_{i\in I}A_i: a(i)\in A_i\forall i\in I\right\}$. If the $A_i$ always is the same, write $A^{I}$ instead of $\prod_{i\in I}A_i$. $\endgroup$ – John_Doe Mar 31 '17 at 7:56
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If $A,B$ are sets then $A^B$ usually denotes the set of functions $B\to A$, and $A\times B:=\{\langle a,b\rangle\mid a\in A\wedge b\in B\}$.

In that context what you write in your title is not true.

However, there is a canonical bijection $M^{\mathbb Z}\to M^{\mathbb Z_{\leq0}}\times M^{\mathbb Z_{>0}}$ prescribed by: $$f\mapsto\langle f\upharpoonleft\mathbb Z_{\leq0},f\upharpoonleft\mathbb Z_{>0}\rangle$$ Its inverse is the map prescribed by:$$\langle g,h\rangle\mapsto g\cup h$$

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  • $\begingroup$ So maybe it is correct that there is an isomorphism? $\endgroup$ – John_Doe Mar 31 '17 at 8:12
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    $\begingroup$ Yes. The map described in my answer is a bijection, which is an isomorphism in the category of sets. $\endgroup$ – drhab Mar 31 '17 at 8:14
  • $\begingroup$ In the map you mean the restriction to $Z_{\leq 0}$ and $Z_{>0}$ right? The inverse function is not totally clear to me yet (the union). $\endgroup$ – John_Doe Mar 31 '17 at 8:16
  • $\begingroup$ Yes, restrictions. If $g:I\to A$ and $h:J\to A$ are functions and $I\cap J=\varnothing$ then $g\cup h$ is a function $I\cup J\to A$. Here $g$ and $h$ are sets of ordered pairs, and so is their union. $\endgroup$ – drhab Mar 31 '17 at 8:20
  • $\begingroup$ Ok. Why do you speak of ordered pairs? I think for example $g$ looks like $(g(1),g(2),g(3),...)$ with $g(1),g(2),...\in Z$. $\endgroup$ – John_Doe Mar 31 '17 at 8:23
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Whether there is exact equality depends on exactly how you construct products and function spaces. In my book, $M^{\Bbb Z}$ and $\prod_{i\in \Bbb Z}M$ are not exactly the same. An element $f\in M^{\Bbb Z}$ is a set of ordered pairs $$ f=\{\ldots,(-1,m_{-1}),(0,m_0),(1,m_1),\ldots\} $$ with $m_i\in M$, while an element $x\in \prod_{i\in \Bbb Z}M$ is an ordered tuple $$ x=(\ldots,m_{-1},m_0,m_1,\ldots). $$ So the two sets aren't equal. Finally, the third set, $M^{\Bbb Z_{\leq 0}}\times M^{\Bbb Z_{>0}}$, consists of pairs of sets of pairs. In other words, an element $y$ looks like this: $$ y=(\{\ldots,(-1,m_{-1}),(0,m_0)\},\{(1,m_1),(2,m_2),\ldots\}) $$ That being said, there are canonical bijections between all these sets.

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