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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?

John_Doe
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    This is correct. – Crostul Mar 31 '17 at 07:40
  • 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 $Z'=Z\cup\left{-\infty,+\infty\right}$? – John_Doe Mar 31 '17 at 07:43
  • How exactly do you define $\prod_{i\in\mathbb{Z}}M_i$ and $M^{\mathbb Z}$? – drhab Mar 31 '17 at 07:46
  • @Crostul I would rather say that it is almost correct. – drhab Mar 31 '17 at 07:50
  • @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 $\prod_{i\in I}A_i$. – John_Doe Mar 31 '17 at 07:56
  • But $A\times B$ is just a set of ordered pairs, right? An indexset $I$ lacks there. That inconsistency is causing trouble. – drhab Mar 31 '17 at 08:02
  • I am not sure I do understand what you mean. – John_Doe Mar 31 '17 at 08:05
  • In short: LHS is set of functions and RHS is set of ordered pairs. Then the sets are not the same (functions are sets of ordered pairs, but are usually not ordered pairs themselves). – drhab Mar 31 '17 at 08:10

2 Answers2

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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:$$\langle g,h\rangle\mapsto g\cup h$$

drhab
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  • So maybe it is correct that there is an isomorphism? – John_Doe Mar 31 '17 at 08:12
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    Yes. The map described in my answer is a bijection, which is an isomorphism in the category of sets. – drhab Mar 31 '17 at 08:14
  • 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). – John_Doe Mar 31 '17 at 08:16
  • 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. – drhab Mar 31 '17 at 08:20
  • Ok. Why do you speak of ordered pairs? I think for example $g$ looks like $(g(1),g(2),g(3),...)$ with $g(1),g(2),...\in Z$. – John_Doe Mar 31 '17 at 08:23
  • Sorry but have to go now. I will come back later. – drhab Mar 31 '17 at 08:32
  • So I guess it is right to say, for example, that $(M^{\mathbb{Z}{\leq 0}}\times M^{\mathbb{Z}{>0}}\times\mathbb{Z})\cup (M^{\mathbb{Z}}\times\left{\pm\infty\right})\simeq (M^{\mathbb{Z}}\times\mathbb{Z})\cup (M^{\mathbb{Z}}\times\left{\pm\infty\right})$ since $M^{\mathbb{Z}{\leq 0}}\times M^{\mathbb{Z}{>0}}\simeq M^{\mathbb{Z}}$. Moreover, we should have the identity (really identity this time) $(M^{\mathbb{Z}}\times\mathbb{Z})\cup (M^{\mathbb{Z}}\times\left{\pm\infty\right})=M^{\mathbb{Z}}\times \mathbb{Z}'$, where $\mathbb{Z}':=\mathbb{Z}\cup\left{\pm\infty\right}$. ? – John_Doe Mar 31 '17 at 09:27
  • Let $Q:=(M^{\mathbb{Z}{\leq 0}}\times M^{\mathbb{Z}{>0}}\times\mathbb{Z})\cup (M^{\mathbb{Z}}\times\left{\pm\infty\right})$ and $P:=M^{\mathbb{Z}}\times\bar{\mathbb{Z}}$. Now, equip $Q$ and $P$ with their associated product topologies $\mathcal{O}_Q$ and $\mathcal{O}_P$. Do we then have that $(Q,\mathcal{O}_Q)\simeq (P,\mathcal{O}_P)$, i.e.. both topological spaces are isomorphic? Maybe this is just via the bijection you have already defined in your answer, since I guess it is continuous with respect to the the product topologies as well as its inverse? – John_Doe Mar 31 '17 at 11:44
  • I speak of ordered pairs because a function is a special set of ordered pairs. See (the answers) on this question. What you say in your comment starting with "So I guess.." is correct. In your last comment we are somehow drifting away from the original question which did not involve any topologies. But yes, my intuition tells me that the spaces will be isomorphic objects in the category of topological spaces also. It goes too far too handle this in a comment. If you want precision then a new question is needed equipped with the tag topology. – drhab Mar 31 '17 at 18:58
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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.

Arthur
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