# Equivalent form of the Whitehead problem

Let $$M$$ be an $$R$$-module and consider the following statement.

$$M$$ is projective whenever the obvious group map $$\tau: \text{Hom}_R(M, R) \to \text{Hom}_R(M, {R}/{I})$$ is surjective for any ideal $$I\subset R$$.

I have read that when $$R = \mathbb{Z}$$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $$M$$ must be free so long as every group extension of $$M$$ by $$\mathbb{Z}$$ is trivial.

I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $$\tau$$ equivalent to the condition that $$\text{Ext}(M, \mathbb{Z}) =0$$? I don't see how the fact that $$\tau$$ is surjective implies that any such extension splits.

I may be missing something obvious, but could someone please explain?

Here is one reference for what I am talking about.

This is incorrect: surjectivity of $$\tau$$ is strictly weaker than $$\operatorname{Ext}(M,\mathbb{Z})=0$$. Specifically, surjectivity of $$\tau$$ is instead equivalent to $$\operatorname{Ext}(M,\mathbb{Z})$$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $$R=\mathbb{Z}$$ and in fact is provably false for $$R=\mathbb{Z}$$. For instance, if $$M=\mathbb{Q}$$, $$\operatorname{Hom}(\mathbb{Q},\mathbb{Z}/I)=0$$ for any ideal $$I\subseteq\mathbb{Z}$$, so $$\tau$$ is trivially always surjective, but $$\mathbb{Q}$$ is not projective.
Here's how you prove that surjectivity of $$\tau$$ is equivalent to $$\operatorname{Ext}(M,\mathbb{Z})$$ being torsion free. Note that for any $$I\subseteq \mathbb{Z}$$, we have a long exact sequence $$0\to \operatorname{Hom}(M,I)\to\operatorname{Hom}(M,\mathbb{Z})\stackrel{\tau}\to\operatorname{Hom}(M,\mathbb{Z}/I)\stackrel{\delta}\to\operatorname{Ext}(M,I)\stackrel{f}\to\operatorname{Ext}(M,\mathbb{Z})\to\operatorname{Ext}(M,\mathbb{Z}/I)\to 0.$$
In particular, we see that $$\tau$$ is surjective iff $$\delta=0$$ iff $$f$$ is injective. Note moreover that if $$I$$ is nontrivial, then $$I=n\mathbb{Z}$$ for some $$n>0$$ and so we can identify $$f$$ with the multiplication by $$n$$ map $$\operatorname{Ext}(M,\mathbb{Z})\to\operatorname{Ext}(M,\mathbb{Z})$$. So, $$\tau$$ is surjective for $$I=n\mathbb{Z}$$ iff $$\operatorname{Ext}(M,\mathbb{Z})$$ has no $$n$$-torsion. We conclude that $$\tau$$ is surjective for all $$I$$ iff $$\operatorname{Ext}(M,\mathbb{Z})$$ is torsion-free.
• Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= \mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like? – CuriousKid7 Jan 20 at 19:05
• Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested. – CuriousKid7 Jan 20 at 19:52