Suppose there is a commutative diagram of vector spaces $A_{i,j}, i,j \ge 1$ with vertical maps $V_{i,j}: A_{i,j} \rightarrow A_{i+1, j}$ and horizontal maps $H_{i,j}: A_{i,j} \rightarrow A_{i, j+1}$. Assume that all vertical maps $V$ are isomorphisms. I care about the direct limit $\lim_{H_{1,j}} A_{1, j}$; note that this is isomorphic to $\lim_{H_{n,j}} A_{n,j}$ for any fixed $n$ or even the diagonal direct limit $\lim_{H_{n+1,n}\circ V_{n,n}} A_{n,n}$ since the vertical maps are isomorphisms.

Suppose that there is another similarly defined commutative diagram $(B_{i,j}, V'_{i,j}, H'_{i,j})$. Suppose that for every $j$, exists $C(j)$ such that for $i, i' \ge C(j)$, $A_{i,j} = A_{i', j} = B_{i,j} = B_{i',j}$ and $A_{i, j+1} = A_{i', j+1} = B_{i,j+1} = B_{i', j+1}$ and $H_{i,j} = H'_{i,j}$; that is, the vector spaces are identical and the horizontal maps agree. However, the vertical isomorphisms $V, V'$ (between identical vector spaces) do not necessarily agree.

Is it true that $\lim_{H_{1,j}} A_{1, j}$ and $\lim_{H'_{1,j}} B_{1, j}$ are isomorphic?

If the diagrams are finite (i.e. $1\le i, j \le C$ for some $C$), then this is true because we can look at the last row of both diagrams, which are the same. In the infinite case, we can write $\lim_{H'_{1,j}} B_{1, j}$ as $\lim_{H_{1,j}\circ \phi_j} A_{1, j}$ for some automorphism $\phi_j$ of $A_{1,j}$. However, it is not true that $\lim_{H_{1,j}\circ \phi_j} A_{1, j}$ and $\lim_{H_{1,j}} A_{1, j}$ are isomorphic in general, even if the direct limit is finite. See my past question: Direct limit isomorphism. The automorphism $\phi_j$ is not the identity because the vertical maps $V, V'$ do not necesssarily agree.


Suppose we have two directed systems $$X:= X_1\to X_2\to X_3\to\dots$$ and $$Y:=Y_1\to Y_2\to Y_3\to\dots$$ such that, for each $n$, the truncations after $n$ terms are isomorphic. I.e., for each $n$ there is a commutative diagram $\require{AMScd}$ \begin{CD} X_1@>>> X_2@>>>\dots@>>> X_n\\ @VV\alpha_{n,1}V @VV\alpha_{n,2}V@.@VV\alpha_{n,n}V\\ Y_1@>>> Y_2@>>>\dots@>>> Y_n \end{CD} with all vertical maps isomorphisms. Note that I'm not assuming any compatibility between the isomorphisms for different $n$.

Then we can construct two diagrams as in the question.

The first diagram has every row equal to $Y$, with identity maps for all vertical maps.

The second diagram has first row equal to $X$, and for $m>1$ the $m$th row is $$X(m):=Y_1\to Y_2\to\dots\to Y_{m-1}\to X_m\to X_{m+1}\to\dots$$ with the map $Y_{m-1}\to X_m$ being the composition $$Y_{m-1}\stackrel{\alpha_{m-1,m-1}^{-1}}{\to}X_{m-1}\to X_m.$$

To describe the vertical arrows, note that for each $m>1$ there is an isomorphism of directed systems $\varphi(m):X\to X(m)$ given by the commutative diagram \begin{CD} X_1@>>> X_2@>>>\dots@>>> X_{m-1}@>>> X_m@>>> X_{m+1}@>>>\dots\\ @VV\alpha_{m-1,1}V @VV\alpha_{m-1,2}V@.@VV\alpha_{m-1,m-1}V @VV\text{id}V@VV\text{id}V@.\\ Y_1@>>> Y_2@>>>\dots@>>> Y_{m-1}@>>> X_m@>>> X_{m+1}@>>>\dots\\ \end{CD} In the second diagram the map from the first row to the second row is $\varphi(2)$, and the map from the $k$th row to the $(k+1)$st is $\varphi(k+1)\varphi(k)^{-1}$ for $k>1$, so that the composition of maps from the first to $k$th row is $\varphi(k)$ for every $k$.

So we just have to find $X$ and $Y$ as above with different direct limits.

Taking vector spaces over a field $k$, let $$Z_1:= k\to0\to0\to0\to\dots$$ $$Z_2:=k\stackrel{\text{id}}{\to}k\to0\to0\to\dots$$ $$Z_3:=k\stackrel{\text{id}}{\to}k\stackrel{\text{id}}{\to}k\to0\to\dots$$ and so on, and $$W:=k\stackrel{\text{id}}{\to}k\stackrel{\text{id}}{\to}k\stackrel{\text{id}}{\to}k\to\dots,$$ and let $X=\bigoplus_i Z_i$ and $Y=\bigoplus_i Z_i\oplus W$. The direct limit of $X$ is zero, but the direct limit of $Y$ is $k$.

  • $\begingroup$ Thank you for your answer. I am bit confused about what you mean "the compositions of vertical maps from the first to mth rows given by $\alpha_{m-1, m-1}^{-1}$ on the first $m−1$ entries, and the identity on the rest"- some more details here would be good. I am also confused why the second diagram is commutative. For example, the left column of the second and third rows involves a square such that the commutativity implies the natural maps $Y_1 \rightarrow X_1 \rightarrow X_2 \rightarrow Y_2$ and $Y_1 \rightarrow Y_1 \rightarrow Y_2$ agree. Why is this? $\endgroup$ – user39598 Apr 3 '17 at 23:09
  • $\begingroup$ I am confused because the first map involves $\alpha_{2,2}$ and $\alpha_{1,1}$ which have nothing to do with each other. $\endgroup$ – user39598 Apr 3 '17 at 23:11
  • $\begingroup$ @user39598 I've added some more details. Let me know if anything is still unclear. $\endgroup$ – Jeremy Rickard Apr 4 '17 at 8:29
  • $\begingroup$ thank you for providing more details. I am also confused why the $X,Y$ you mention at the very end satisfy the necessary conditions; it seems to me that $X_1 = k^3$ but $Y_1 = k^4$ so that $X_1 \not \cong Y_1$, which is a necessary condition for the construction. $\endgroup$ – user39598 Apr 4 '17 at 17:54
  • $\begingroup$ @user39598 I think you missed the "and so on". $\endgroup$ – Jeremy Rickard Apr 4 '17 at 18:03

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