Weakest possible condition for $\mathrm{Hom}(H,\mathbb{Z}) \otimes G \cong \mathrm{Hom}(H,G)$

Question

Let's say that we have two abelian groups, $H$ and $G$. We can view them as $\mathbb{Z}$-modules, so it makes sense to tensor them over $\mathbb{Z}$. We also get a somehow obvious map:

$$\mathrm{Hom}(H,\mathbb{Z}) \otimes G \longrightarrow \mathrm{Hom}(H,G); f \otimes g \mapsto (h \mapsto f(h) \cdot g)$$

It is well-known and relatively easy to prove, that if both $H$ and $G$ are free and finitely generated abelian, then the above map is an isomorphism:

$$\mathrm{Hom}(H,\mathbb{Z}) \otimes G \cong \mathrm{Hom}(H,G)$$

This isomorphism also works for arbitrary free modules with finite rank over a ring. But if we take the ring to be $\mathbb{Z}$, it is often the case that stronger statements hold. Explicitely I am wondering, if we can loosen some constraints on $H$ or $G$ if we take the ring to be $\mathbb{Z}$.

My suspicion is that this isomorphism still holds, if we only require $H$ to be free and finitely generated, and $G$ to be any arbitrary abelian group. Because in this case we have an isomorphism $H \cong \mathrm{Hom}(H, \mathbb{Z})$.

Does this work, and if yes, is it possible to even require less?

For some context, I am interested in some minimal conditions on the homology groups of spaces or generally chain complexes, such that I can formulate a statement of the form $H^n(X;R) \cong H^n(X;\mathbb{Z})\otimes R$ for some ring of coefficients $R$.

Answer 1

If $H$ is free on $n$ generators, then so is $\operatorname{Hom}(H,\mathbb Z),$ and $H\cong\operatorname{Hom}(H,\mathbb Z),$ so you really want $H\otimes G\cong \operatorname{Hom}(H,G).$

Let $X=\{x_1,\dots,x_n\}$ be the generators.

Every element of $H\otimes G$ can be written uniquely as a sum: $$\sum_i x_i\otimes g_i$$ for some tuple $(g_1,\dots,g_n).$ We can easily show $H\otimes G\cong G^n.$

Similarly, we can see an $f:H\to G$ is uniquely determined by $h(x_i)=g_i$ for $i=1,\dots,n$ and we can show $\operatorname{Hom}(H,G)=G^n.$

If $H$ is free but with an infinite cardinal $\alpha$ of generators, then it is still true that $\operatorname{Hom}(H,G)=G^\alpha,$ but $H\otimes G$ in that case is isomorphic to a subgroup of $G^\alpha$ where only finitely many of the $g_i$ are non-zero.

Thus, the cardinality of the groups are often (always?) different. For example, if $1<|G|\leq \aleph_0$ and $\alpha=\aleph_0,$ you have $H\otimes G$ with the cardinality $\aleph_0$ and $|G^\alpha|$ the cardinality of the continuum.

---

If $f_i:H\to \mathbb Z$ are defined by $$f_i(x_j)=\delta_{ij}=\begin{cases}1&i=j\\0&i\neq j\end{cases}$$ then the $f_i$ are a basis for $\operatorname{Hom}(H,\mathbb Z)$ and every element of $\operatorname{Hom}(H,\mathbb Z)\otimes G$ Can be written uniquely as: $$\sum_i f_i\otimes g_i$$ for some $(g_1,\dots,g_n).$

Then the image of $f_i\otimes g_i$ under your homomorphism is the map that sends the basis element:

$$x_j\mapsto \delta_{i,j}g_i.$$

Show that this sends the representation of $G^n$ on the left to the representation of $G^n$ on the right.

---

A trivial extension: If $H=H_1\oplus H_2$ where $H_1$ is free with finite generators and every element of $H_2$ has finite order, and $G$ is a divisible group, then $\operatorname{Hom}(H,\mathbb Z)\cong\operatorname{Hom}(H_1,\mathbb Z)\cong H_1$ and $H\otimes G=H_1\otimes G.$

Answer 2

It suffices for either $G$ or $H$ to be finitely generated and free, and then the other can be completely arbitrary. Note first that if either $G$ or $H$ is $\mathbb{Z}$, you can easily verify your map is an isomorphism (since $\mathbb{Z}\otimes A\cong A$ and $\operatorname{Hom}(\mathbb{Z},A)\cong A$). Now observe that the functors on both sides as well as your natural transformation between them preserve finite direct sums. A finitely generated free module is just a finite direct sum of copies of $\mathbb{Z}$, so it follows your natural transformation is an isomorphism in that case.

(Nothing about $\mathbb{Z}$ here is special; this also works for modules over any ring. In fact, more generally, this shows that it would suffice for either $G$ or $H$ to be a direct summand of a finitely generated free module. Over $\mathbb{Z}$ this is the same as just being a finitely generated free module, but in general it is weaker, and is equivalent to being finitely generated and projective.)