# Projective resolution of a filtered colimit

In Cartan and Eilenberg's Homological Algebra, they claim the following

V9.5* If $$A = \lim\limits_\longrightarrow A_\alpha$$ then there exist projective resolutions $$X_\alpha$$ of $$A_\alpha$$ forming a direct system such that $$X = \lim\limits_\longrightarrow X_\alpha$$ is a projective resolution of $$A$$.

I'm assuming they intend for a direct system to mean, as usual, that the index set is directed and that there are maps so that we have maps $$\phi_{\alpha \beta}: A_\alpha \rightarrow A_\beta$$ for any $$\alpha \leq \beta$$ respecting transitivity and being the identity in case $$\alpha = \beta$$.

I really want to believe this lemma, but I'm having trouble with their brief proof. This result surprises me because it would immediately imply that Tor commutes with filtered colimits, and the proofs of that known to me are both a good deal more involved than this.

The idea of the proof is simple. They construct the projective resolutions $$X_\alpha$$ simultaneously, and one step at a time just like any projective resolution. At each step they claim that the partially constructed resolutions form a directed system with maps induced by the previous step. This is where I get lost.

Consider the first step. Cartan and Eilenberg start the resolutions off with $$F_\alpha \rightarrow A_\alpha \rightarrow 0$$. They claim that the maps $$A_\alpha \rightarrow A_{\alpha'}$$ induce maps $$F_\alpha \rightarrow F_{\alpha'}$$.

But why, I wonder, should it be the case that $$F_\alpha \xrightarrow{\phi_{aa'}} F_{\alpha'} \xrightarrow{\phi_{a'a''}} F_{\alpha''}$$ is the same map as $$F_\alpha \xrightarrow{\phi_{aa''}} F_{\alpha''}$$??. The induced maps come from nonuniversal lifts, so, as far as I can tell, we can only say that $$\phi_{a a'}\phi_{a' a''} - \phi_{aa''}$$ maps to $$\ker(F_{\alpha''} \rightarrow A_{\alpha''})$$, but we haven't actually constructed a directed system of modules $$F_\alpha$$.

Is there something I'm missing here? Is their argument perfectly fine as is and I'm just being dense? If not, is there a fix to it?

Let $$X^{(Y)}$$ mean a direct sum of copies of $$X$$, indexed by the set $$Y$$. I'll assume left $$R$$-modules are being dealt with.

The trick, given a module $$M$$, is to set $$F(M)=R^{(M)}$$, with the map $$\pi_M\colon F(M)\to M$$ defined by $$\pi_M\Bigl(\sum_{x\in M}r_xe_x\Bigr)=\sum_{x\in M}r_xx$$ where $$e_x$$ is the vector having $$1$$ at the $$x$$-component and $$0$$ elsewhere. In other words, we define $$\pi_M$$ on the basis by sending $$e_x$$ to $$x$$.

If $$f\colon M\to N$$ is a module homomorphism, then $$f$$ induces a module homomorphism $$\tilde{f}\colon F(M)\to F(N)$$ by $$\tilde{f}\Bigl(\sum_{x\in M}r_xe_x\Bigr)=\sum_{x\in M}r_xe_{f(x)}$$ In other words, we define $$\tilde{f}$$ on the basis by sending $$e_x$$ to $$e_{f(x)}$$.

Now $$\pi_N\tilde{f}(e_x)=\pi_N(e_{f(x)}=f(x)$$ and $$f\pi_N(e_x)=f(x)$$. Therefore $$\pi_N\tilde{f}=f\pi_M$$ because both homomorphisms coincide on a basis of $$F(M)$$.

In a more abstract way, this is the adjunction “free module”–“forgetful functor”.

• Sorry if I wasn't clear in my question, but what I'm asking is specifically why/ how the $F(M)$ are supposed to form a direct system with the right limit. I agree that those are the only reasonable maps to consider. In your notation, we get that $\pi_N f_{MN}^~ f_{KM}^~ = \pi_N f_{KN^~}$, but what we actually need for a directed system is $f_{MN}^~ f_{KM}^~ = f_{KN^~}$. Even working with index set the natural numbers, I don't see how to ensure that. – Badam Baplan Jun 27 '19 at 16:35
• @BadamBaplan It's functorial. – egreg Jun 27 '19 at 16:48
• of course it is! thanks for bearing with me :) For some reason at 4am I was just thinking of the free modules as projective. – Badam Baplan Jun 27 '19 at 17:10