# inverse limit of isomorphic vector spaces

Let $$\cdots \rightarrow A_{n+1}\rightarrow^{f_{n+1}} A_n \rightarrow^{f_{n}} A_{n-1}\rightarrow \cdots$$ be an inverse system of finite dimensional vector spaces with the property that the $A_i$ are 'eventually constant', i.e., there is an $m$ such that the maps $f_i$ are isomorphisms for every $i\ge m$. Does it follow that $$\lim_{\leftarrow} A_n \simeq A_m?$$

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Yes............ –  Robin Chapman Nov 11 '10 at 18:36
could you please indicate why? –  Laurent S Nov 11 '10 at 18:53
Use the universal property for inverse limits with appropriate projection maps. –  Nuno Nov 11 '10 at 19:02
Ok, thanks Nuno. I'll try to use that. If anyone wants to fill in the details, feel free to do so. –  Laurent S Nov 11 '10 at 19:25

Remember that if your directed set has a maximum, then the inverse limit is just the maximum; here this is "essentially" what you have, since after a certain point you are not really getting anything new: you can identify all $A_i$ with $i\geq m$, collapsing the "left tail" of your inverse system into a terminating one, so the inverse system is just that maximum.

In fact, this works whether you are working with finite dimensional vector spaces or any kind of structure.

But of course, "essentially" is not the same as "exactly." So you use the universal property. I claim that $A_m$ has the desired property. Let $f_{ij}$ with $i\geq j$ be defined as the identity if $i=j$, and as the composition $f_i\circ f_{i-1}\circ\cdots\circ f_j$ if $i\gt j$. Then the projection maps $\pi_j\colon A_m\to A_j$ are defined by $\pi_j=f_{mj}$ if $j\leq m$, and as $\pi_j = f_{jm}^{-1}$ if $j\gt m$; note that this makes sense since $f_{jm}$ is a composition of isomorphisms when $j\gt m$, so it is itself an isomorphism and has an inverse. Note that for any $i\gt j$ we have $f_{ij}\circ \pi_j = \pi_i$, since you have $f_{ij}\circ f_{jk} = f_{ik}$ whenever $i\geq j \geq k$.

Now let $P$ be any object together with maps $p_j\colon P\to A_j$ such that for all $i\gt j$ you have $f_{ij}\circ p_i = p_j$. Then $p=p_m\colon P\to A_m$ has the appropriate property, that is, $p_j = \pi_j\circ p$ for all $j$: if $j\leq m$, then this because $\pi_j = f_{mj}$, so $\pi_j\circ p = f_{mj}\circ p_m = p_j$, and if $j\gt m$ then $f_{jm}\circ p_j = p_m$, so pre-composing with $f_{jm}^{-1}$ we get $p_j = (f_{jm})^{-1}\circ p_m = \pi_j\circ p$. The map is unique: if $f\colon P\to A_m$ has the same property, then $f=f_m = p$.

Thus, $(A_m, \{\pi_j\})$ has the desired universal property, and so "is" the inverse limit.

You'll note that the fact that we are dealing with finite dimensional vector spaces is completely immaterial; what matters is the properties of the functions in play. This is typical of universal constructions.

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Nice answer Arturo. I was too lazy to post one by myself. –  Nuno Nov 11 '10 at 20:08
I agree, this is a really great answer! Thanks! –  Laurent S Nov 11 '10 at 21:06