# Direct limit isomorphism

Suppose I have a directed system $(V_i, \phi_i: V_i \rightarrow V_{i+1})$, say of vector spaces $V_i$. Let $\psi_i: V_i \rightarrow V_i$ be isomorphisms. I can construct the related directed system $(V_i, \phi_i\circ \psi_i: V_i \rightarrow V_{i+1})$. Is it true that the direct limits $\lim_{\phi_i} V_i$ and $\lim_{\phi_i\circ \psi_i} V_i$ are isomorphic? It seems to me that you cannot directly use $\psi_i$ to get isomorphism of the directed system but maybe the limits are still isomorphic.

Edit: As explained in Jeremy's answer below, this is false in general. What about if $\phi, \psi$ commute? ie $\phi_i \psi_i = \psi_{i+1}\phi_i$. In this case, $\psi$ induces an isomorphism of $\lim_{\phi_i} V_i$. Does this stronger assumption imply that $\lim_{\phi_i} V_i$ and $\lim_{\phi_i\circ \psi_i} V_i$ are isomorphic? In Jeremy's example, $\phi, \psi$ do not commute.

Take $V_i=k^2$ for all $i$, with $\phi_i$ given by the matrix $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ for all $i$. Then $\varinjlim_{\phi_i}V_i$ is one dimensional.
But now take $\psi_i$ to be the isomorphism given by the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ for all $i$. Then $\varinjlim_{\phi_i\circ\psi_i}V_i$ is zero.
The answer to the supplementary question about the case where $\phi_i\psi_i=\psi_{i+1}\phi_i$ is that in this case the directed systems are isomorphic (via $\psi_i^i:V_i\to V_i$), and therefore the direct limits are isomorphic.
• What about when $\phi_i$ and $\psi_i$ commute? ie. $\phi_i \psi_i = \psi_{i+1} \phi_i$? I edited my question to incorporate this question as well. – user39598 Mar 11 '17 at 18:40
• Can you provide details about the supplementary question? I don't see why the systems are isomorphic. If $\phi, \psi$ commute, then of course $\psi_i$ is an isomorphism of the directed system $(V_i, \phi_i : V_i \rightarrow V_{i+1})$ and hence we get an automorphism $\psi$ of $\lim_{\phi_i} V_i$. However, it is not clear to me that this implies that $(V_i, \phi_i : V_i \rightarrow V_{i+1})$ , $(V_i, \phi_i \circ \psi_i: V_i \rightarrow V_{i+1})$ are isomorphic or that their direct limits are isomorphic. – user39598 Mar 13 '17 at 7:09
• @user39598 I gave the isomorphism in my answer. They're isomorphic via the $i$th iterate $\psi_i^i:V_i\to V_i$. This is an isomorphism for each $i$, and $\psi_{i+1}^{i+1}\circ\phi_i=(\phi_i\circ\psi_i)\circ\psi_i^i$, so the relevant squares involving $V_i$ and $V_{i+1}$ commute. – Jeremy Rickard Mar 13 '17 at 8:48
• Ok, thank you! I misread the isomorphism and didn't notice that you are considering the $i$th iterate. – user39598 Mar 13 '17 at 19:18