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Let $A$ be a finite dimensional vector space, $\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, and let $\phi_i:A\to A_i$ be linear maps inducing a map $\phi:A\to\lim\limits_{\leftarrow} A_i$. If $\phi$ is an injection (resp. surjection), does it follow that $\phi_i$ is an injection (resp. surjection) for some $i\gg 0$?

My guess is that it does, since the dimensions $\dim ker \phi_i$ and $\dim coker \phi_i$ should be non-increasing sequences.

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you need to rename $f$ and the $f_i$ –  Plop Nov 14 '10 at 16:03
    
fixed, thanks for the correction. –  Laurent S Nov 14 '10 at 16:05

2 Answers 2

up vote 3 down vote accepted

If $\mathbf{v}\in A$ is in the kernel of $\phi_i$, then for all $j\leq i$ you have $\phi_j(\mathbf{v}) = f_{ij}\phi_j(\mathbf{v}) = f_{ij}(\mathbf{0}) = \mathbf{0}$ (where $f_{ij}\colon A_i\to A_j$ is the map obtained by composing the $f_k$). Therefore, $\mathrm{ker}(\phi_i)\subseteq \mathrm{ker}(\phi_j)$ for all $i\geq j$. Hence you have a descending sequence \begin{equation*} \cdots \subseteq \mathrm{ker}(\phi_{i+1})\subseteq \mathrm{ker}(\phi_i)\subseteq \cdots \subseteq \mathrm{ker}(\phi_0). \end{equation*} Since $\dim(A)\lt\infty$, the sequence must stabilize, so there exists $n_0$ such that for all $n\geq n_0$, $\mathrm{ker}(\phi_n)=\mathrm{ker}(\phi_{n_0})$. Let $\mathbf{N}$ be that subspace of $A$.

If $\mathbf{v}\in\mathbf{N}$, then $\phi_i(\mathbf{v})=\mathbf{0}$ for all $i$, hence $\phi(\mathbf{v})=\mathbf{0}$, so $\mathbf{N}\subseteq \mathrm{ker}(\phi)$. Conversely, anything in $\mathrm{ker}(\phi)$ lies in the intersection of all kernels of the $\phi_i$. Thus, $\mathbf{N}=\mathrm{ker}(\phi)$.

In conclusion, $\phi$ is one to one if and only if the $\phi_i$ are eventually one-to-one.

For surjection, you are going to be out of luck. Consider the system with $A_i = \mathbf{F}^2$ for all $i$, $f_i\colon\mathbf{F}^2\to\mathbf{F}^2$ given by $f_i(a,b)=(a,0)$. I claim that the inverse limit of this system is $\mathbf{V}=\mathbf{F}$ with structure maps $\mathfrak{f}_i\colon\mathbf{V}\to\mathbf{F}^2$ given by $\mathfrak{f}(a) = (a,0)$. First, note that this is a consistent system of maps from $\mathbf{V}$ to the $A_i$. Next, suppose that $\mathbf{W}$ is any vector space and that you have maps $g_i\colon\mathbf{W}\to\mathbf{F}^2$ for all $i$, with the property that for all $i\geq j$, $g_j(\mathbf{w}) = f_{ij}(g_i(\mathbf{w}))$. If $g_i(\mathbf{w}) = (\mathbf{w}_{1i},\mathbf{w}_{2i})$, then this means that $g_j(\mathbf{w}) = (\mathbf{w}_{1i}, 0)$. In particular, for all $j$ we have that $\mathrm{Im}(g_j)$ is contained in the subspace $\{(x,0)\mid x\in\mathbf{F}\}$ of $A_j$, and since the $f_i$ are isomorphisms when restricted to those subspaces, then it follows that for each $\mathbf{w}\in\mathbf{W}$ there is an $a_{\mathbf{w}}\in\mathbf{F}$ such that $g_j(\mathbf{w}) = (a_{\mathbf{w}},0)$ for all $j$. Hence, we have a map $\mathfrak{g}\colon\mathbf{W}\to \mathbf{V}$ with $\mathfrak{f}_j\circ\mathfrak{g}=g_j$ for all $j$, namely $\mathfrak{g}(\mathbf{w}) = a_{\mathbf{w}}$, and the map is clearly unique. Thus, $(\mathbf{V},\mathfrak{f}_i)$ is the inverse limit of the system. Then $\mathbf{V}$ itself with the structure maps gives a counterexample: the induced map to the inverse limit is onto (in fact, it is the identity), but the structure maps are never onto, so it is false that $\phi_i$ is onto for sufficiently large $i$.

Added: However, if your structure maps $f_i$ are eventually surjective, then your conclusion will follow. To see this, assume that $f_i$ is surjective for all $i\geq N$; I claim that the inverse limit maps $\mathfrak{f}_j\colon\lim\limits_{\leftarrow}A_i\to A_j$ are surjective for all $j\geq N$. Fix $j\geq N$, and let $\mathbf{v}\in A_j$. Let $\mathbf{v}_{j+1}\in A_{j+1}$ be any preimage of $\mathbf{v}$, and inductively define $\mathbf{v}_{j+k+1}$ to be any preimage of $\mathbf{v}_{j+k}$. Then define maps $g_i\colon\mathbf{F}\to A_i$ by letting $g_i(1) = \mathbf{v}_{j+m}$ if $i=j+m\geq N$, and $g_i(1)=f_{ji}(\mathbf{v})$ if $i\lt N$. This induces a map $\mathfrak{g}\colon\mathbf{F}\to\lim\limits_{\leftarrow}A_i$, and $g_i = \mathfrak{f}_i\circ\mathfrak{g}$. In particular, $\mathbf{v}\in\mathrm{Im}(\mathfrak{f}_j)$. Thus, $\mathfrak{f}_j$ is onto for all $j\geq N$. Then, if you have $\phi\colon A\to A_i$ as in your statement surjective, then $\phi_j = \mathfrak{f}_j\circ\phi\colon A\to A_j$ is a composition of surjective functions for alL $j\geq N$, so the $\phi_j$ are eventually surjective, as desired.

Added 2: To see how I came up with the example, notice that if you let $B_{i-1} = \mathrm{Im}(f_i)$, then you get an inverse system with the $B_i$ instead of the $A_i$; the inverse limit of this system is isomorphic to the inverse limit of the original system (try to prove this). This is because a consistent system of homomorphism $\phi_i\colon A\to A_i$ has to have $\mathrm{Im}(\phi_i)\subseteq B_i$. So what really matters is what is going on in the images of the structure maps $f_i$, hence surjectivity into the limit depends only on how your maps $f_i$ relate to the $B_i$, not how they relate to the $A_i$. That is why if the maps $f_i$ are eventually surjective you do get the implication.

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Like you said, we have $\ker\phi_{i+1}\subseteq \ker\phi_{i}$. Hence the decreasing sequence of kernels either stabilizes at some nontrivial subspace of $A$, in which case $\phi$ is not injective, or it decreases to $0$. But it has to decrease to $0$ after finitely many $i$, so after that, $\phi_i$ is injective.

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