Define \(\displaystyle{F}:{H}{o}{m}{\left({R},{M}\right)}\to{M}\) by \(\displaystyle{F}{\left(\phi\right)}=\phi{\left({1}\right)}\in{M}\)

First, we proved that F is an R-module homomorphism.

Let \(\displaystyle\phi,\rho\in{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{r}\in{R}\). Then,

\(\displaystyle{F}{\left({r}\phi-\rho\right)}={\left({r}\phi-\rho\right)}{\left({1}\right)}={r}\phi{\left({1}\right)}-\rho{\left({1}\right)}={r}{F}{\left({r}\right)}-{F}{\left({r}\right)}\)

Hence, F is an R-module homomorphism.

Now, to prove that it is an isomorphic R-module, we must prove that F is bijective.

Suppose \(\displaystyle{F}{\left(\phi\right)}={0}\) for some \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}\). Then, \(\displaystyle\phi{\left({1}\right)}={0}.\)

For any \(\displaystyle{r}\in{R},\phi{\left({r}\right)}={r}\phi{\left({1}\right)}={r}\cdot{0}={0}\). Hence, \(\displaystyle\phi={0}\). Thus, F is injective.

Now, suppose \(\displaystyle{m}\in{M}\).

Define a map \(\displaystyle\phi:{R}\to{M}\) by \(\displaystyle\phi{\left({r}\right)}={r}{m}\) for any \(\displaystyle{r}\in{R}.\)

Let \(\displaystyle{r},{s},{t}\in{R}.\)

\(\displaystyle\Rightarrow\phi{\left({r}{s}-{t}\right)}={\left({r}{s}-{t}\right)}{m}={r}{\left({s}{m}\right)}-{t}{m}={r}\phi{\left({s}\right)}-\phi{\left({t}\right)}\)

Hence, \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}.\)

Futher, F \(\displaystyle{\left(\phi\right)}=\phi{\left({1}\right)}={m}.\)

Thus, F is surjective.

F is homomorphism and bijective. Therefore, \(\displaystyle{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{M}\) are isomorphic R-module.

First, we proved that F is an R-module homomorphism.

Let \(\displaystyle\phi,\rho\in{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{r}\in{R}\). Then,

\(\displaystyle{F}{\left({r}\phi-\rho\right)}={\left({r}\phi-\rho\right)}{\left({1}\right)}={r}\phi{\left({1}\right)}-\rho{\left({1}\right)}={r}{F}{\left({r}\right)}-{F}{\left({r}\right)}\)

Hence, F is an R-module homomorphism.

Now, to prove that it is an isomorphic R-module, we must prove that F is bijective.

Suppose \(\displaystyle{F}{\left(\phi\right)}={0}\) for some \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}\). Then, \(\displaystyle\phi{\left({1}\right)}={0}.\)

For any \(\displaystyle{r}\in{R},\phi{\left({r}\right)}={r}\phi{\left({1}\right)}={r}\cdot{0}={0}\). Hence, \(\displaystyle\phi={0}\). Thus, F is injective.

Now, suppose \(\displaystyle{m}\in{M}\).

Define a map \(\displaystyle\phi:{R}\to{M}\) by \(\displaystyle\phi{\left({r}\right)}={r}{m}\) for any \(\displaystyle{r}\in{R}.\)

Let \(\displaystyle{r},{s},{t}\in{R}.\)

\(\displaystyle\Rightarrow\phi{\left({r}{s}-{t}\right)}={\left({r}{s}-{t}\right)}{m}={r}{\left({s}{m}\right)}-{t}{m}={r}\phi{\left({s}\right)}-\phi{\left({t}\right)}\)

Hence, \(\displaystyle\phi\in{H}{o}{m}{\left({R},{M}\right)}.\)

Futher, F \(\displaystyle{\left(\phi\right)}=\phi{\left({1}\right)}={m}.\)

Thus, F is surjective.

F is homomorphism and bijective. Therefore, \(\displaystyle{H}{o}{m}{\left({R},{M}\right)}{\quad\text{and}\quad}{M}\) are isomorphic R-module.