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

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 $$R$$.

• If $G$ is finite and $H$ is free on $n$ elements, then both groups have size $|G|^n,$ so if you have an inclusion in one direction, it is true for finite $G.$ Commented Aug 19, 2021 at 17:58
• If at least one among $H$ and $G$ is finitely generated and free, then the map is bijective. Commented Aug 19, 2021 at 21:33

## 2 Answers

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 $$H\otimes G=H_1\otimes G.$$

• Thanks a lot! Just to be clear, this means that if $H$ is free and finitely generated abelian, then this map is an isomorphism, regardless of $G$? Commented Aug 19, 2021 at 20:54
• Yeah, I skipped the specific map, and only answered the title. Sorry, will think about it more - Commented Aug 19, 2021 at 21:09

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.)

• Though for a general ring you only need one of the modules to be finitely generated projective, not necessarily free. Commented Aug 19, 2021 at 22:56
• Good point, I've added a bit on that. Commented Aug 19, 2021 at 23:15