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The following notations is inherited from Homology. Suppose we have an exact sequence $$0\rightarrow B_{n}\rightarrow Z_{n}\rightarrow H_{n}(C)\rightarrow 0$$ where $B_{n}$ is the nth boundaries, and $Z_{n}$ is the nth cycles, and $H_{n}(C)$ the homology group of chain complex. The map between $B_{n}$ and $Z_{n}$ is the injective map and the map between $Z_{n}$ and $H_{n}(C)$ is the quotient map induced by boundary map. The proof of Universial Coefficient Theorems implies that $$0\rightarrow \text{Hom}(H_{n}(C),G)\rightarrow \text{Hom}(Z_{n},G)\rightarrow \text{Hom}(B_{n},G)\rightarrow \text{Ext}(H_{n}(C),G)\rightarrow 0 $$ is exact.

So $$\text{Hom}(Z_{n},G)\rightarrow \text{Hom}(B_{n},G)\rightarrow 0$$ is generally not exact at $\text{Hom}(B_{n},G)$,i.e. the injective map from $B_{n}$ to $Z_{n}$ does not induce surjective map from $\text{Hom}(Z_{n},G)$ to $\text{Hom}(B_{n},G)$. However I think I can prove this( of course somewhere wrong ).

Here is my proof.

For any $f\in \text{Hom}(B_{n},G)$,the main thing here is about whether we can complete the commutative diagram below with a map $g\in \text{Hom}(Z_{n},G)$.

enter image description here

But since $Z_{n}$ is free abelian group, I can write it as $$Z_{n}=i(B_{n})\oplus W$$ for some complementary subspace $W$. We define a map $g$ as $g(x)=f(x)$ for $x\in i(B_{n})$ and $g(x)=0, x\in W$ So I get a surjective from $\text{Hom}(Z_{n},G)$ to $\text{Hom}(B_{n},G)$ taking $f$ to $g$, i.e. for any $G$ and free abelian group $B_{n},Z_{n}$, $G$ is injective. I know this is wrong since it contradicts the UCT, but I can not see why. Thank you for any help.

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Great question! I think your mistake is that you slightly mixed up the properties of injective and projective modules.

A free abelian group doesn't have a complement to every its subgroup: take $2\mathbb{Z} \subset \mathbb{Z}$ for instance. To answer the question in the title, the category of free abelian groups has no non-trivial injective objects as illustrated by my example above^ (which works just as well for arbitrary free abelian groups). Further can be said, in fact: any injective object in the (abelian!) category of abelian groups can't be free, since these are precisely the so-called 'divisible' abelian groups (i.e. those for which multiplication by any integer is surjective) - you can look up Baer's criterion if you're interested.

I hope this helps! :)


Edit: just to see what the issue at play is, if you look at the diagram in your post, try and set $G=B_n=Z_n=\mathbb{Z}$, let $f$ be the identity and the injection $B_n \hookrightarrow Z_n$ multiplication by two :)

A very similar diagram occurs for instance when computing the homology of the real projective plane $\mathbb{R} P^2$.

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    $\begingroup$ Thank you. Very helpful answer. I think I mixed up some properties of free abelian group and vector space. $\endgroup$
    – user914799
    Commented Nov 21, 2022 at 7:55

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