Question on Hom set for fields in the context of universal coefficient theorem

Question

My question may be a stupid one or is more about the way of writing things, but I cannot find the answer right now, by myself or via the internet.

I encountered this problem in the context of the Universal coefficient theorem for field-valued relative cohomology. Indeed from the fact that it is field-valued, we have the isomorphism: \begin{equation} H^k(\mathcal{M}, \mathcal{N}; \mathbb{Z}_2) \simeq \text{Hom}_{\mathbb{Z}_2}\left( H_k(\mathcal{M}, \mathcal{N}; \mathbb{Z}_2), \mathbb{Z}_2 \right) \end{equation} Now, what happens if $H_k(\mathcal{M}, \mathcal{N}; \mathbb{Z}_2) = 0$? The $\text{Hom}_{\mathbb{Z}_2}(-,\mathbb{Z}_2)$ functor should act on $\mathbb{Z}_2$-valued objects, but for me, $0$ is not such an object... Or maybe I am missing something obvious?

Maybe we write $0$ instead of $0\times \mathbb{Z}_2$, and in that case $\text{Hom}_{\mathbb{Z}_2}(0\times \mathbb{Z}_2,\mathbb{Z}_2)\simeq \mathbb{Z}_2$?

Answer 1

The $\text{Hom}_{\mathbb{Z}_2}(-,\mathbb{Z}_2)$ functor should act on $\mathbb{Z}_2$-valued objects, but for me, $0$ is not such an object...

Why not? In fact given any ring $R$, we have the unique (up to isomorphism) zero $R$-module, defined as

$$0=\{\theta\}$$ $$\theta + \theta := \theta$$ $$r\cdot\theta:=\theta\text{ for any }r\in R$$

Not that we have much choice for the addition and scalar multiplication.

And then $\text{Hom}_R(0, M)\simeq 0$ for any $R$-module $M$.