# Is there an injective object in the category of all free abelian group?

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 $$g\in \text{Hom}(Z_{n},G)$$.

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.

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.
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 $$\mathbb{R} P^2$$.