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I am trying to understand the proof of a Lemma from Greenberg's Algebraic Topology. It says:

Let $I$ be a directed set and $M_i$ a family of modules. Suppose that for every $i\in I$, we have $M_i=N_i\oplus P_i$. If $i\leq i'$, then we have a descomposition of the homomorphism $\phi_{i',i}:M_i\longrightarrow M_{i'}$, that is, $\phi_{i',i}=\psi_{i',i}+\rho_{i',i}$. Taking $N=\lim\limits_{\to} N_i$, $\psi_{i',i}:N_i\longrightarrow N$; and $P=\lim\limits_{\to} P_i$, $\rho_{i',i}:P_i\longrightarrow P$. Using the definition of limits, we have that, for $M=\bigcup_i \phi_i(M_i)$, there exists $\psi:N\longrightarrow M$ and $\rho:P\longrightarrow M$ verifying $\psi\psi_i=\phi_i$ on $N_i$ and $\rho\rho_i=\phi_i$ on $P_i$. In that case, the map $\psi\oplus \rho$ is an isomorphism.

The proof is:

Given $x\in M$, choose $x_i\in M_i$ such that $x=\phi_i(x_i)$. $x_i=y_i+z_i$ because $M_i=N_i\oplus P_i$. Construct the inverse $\theta(x)=(\psi(y_i),\rho(z_i)\in N\oplus P$. My problem is verifying that $\theta(x)$ does not depend on the choice of $x_i$.

EDIT

There was a mistake. The inverse $\theta$ of the function $\psi\oplus\rho$ should be: $\theta(x)=(\psi_i(y_i),\rho_i(z_i))$.

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Let $\phi_i(x_i) = x = \phi_j(x_j)$. Since $I$ is a direct system, there's $k \in I$ such that $i,j \le k$. Then $\phi_k \circ \phi_{k,i}(x_i) = \phi_i(x_i) = \phi_j(x_j) =\phi_k \circ \phi_{k,j}(x_j) $.

But $\psi \circ \psi_i = \phi_i$ on $N_i$ and $\rho \circ \rho_i = \phi_i$ on $P_i$. Thus, $\psi \circ \psi_i (y_i) = \psi \circ \psi_k \circ \psi_{k,i} (y_i) = \psi \circ \psi_k \circ \psi_{k,j} (y_j) = \psi \circ \psi_j (y_j)$.

The argument is analogous for $\rho \circ \rho_i$.

I'm assuming that you've already proved that $M$ is the direct limit of the $M_i$'s.

Now use the fact that direct limits takes exact sequences into exact sequences. Actually we just need that direct limits preserve kernels. Note that $\psi: N \rightarrow M$ comes from $\delta_i : N_i \rightarrow N_i \oplus P_i$ and this imply that $\psi$ is injective. Then we have that $\psi_i(y_i) = \psi_j(y_j)$. The same holds for $\rho_i(z_i)$ and $\rho_j(z_j)$.

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  • $\begingroup$ You are right, there was a mistake, but not that one. $\endgroup$ – user203327 Apr 10 '16 at 9:39
  • $\begingroup$ I just noticed, The way I defined $\theta$ was assigning $M$ to $M$. Now It's adapted to the correct definition. $\endgroup$ – Garcez Apr 10 '16 at 21:33

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