Question related to "example of free module with bases of different cardinality" In this Planet Math article, it shows an example where a free module has bases with different cardinality.
I am having some problems with show why $\phi$ is an $R$-module homomorphism.
 A: I agree with D_S's comment that this is best done on your own, and probably first in the case $R^2 \cong R$ instead of the general version you're looking at. But if it helps, I'll give the full proof of the requested fact. Here's the setup from the linked article:
$k$ is a field, $V$ is an infinite-dimensional $k$-vector space with basis $\{e_i\}_{i \in I}$, $R = \operatorname{End}_k(V)$, and $J$ is a set such that $\lvert J \rvert \leq \lvert I \rvert$.
Then we pick any bijection $\alpha : I \to J \times I$, and $\pi_1 : J \times I \to J$ and $\pi_2: J \times I \to I$ denote the canonical projections. Now $\phi : R^J \to R$ is defined as follows:
For each $f \in R^J$, $\phi(f) : V \to V$ is the unique endomorphism such that
$$\phi(f)(e_i) = f(\pi_1(\alpha(i)))(e_{\pi_2(\alpha(i))})$$
for all $i \in I$.
To show that $\phi$ is an $R$-module homomorphism, let $f, g \in R^J$ and $\lambda \in R$ be arbitrary. Recall that the $R$-module structure on $R^J$ is defined such that $(f + \lambda g)(j) = f(j) + \lambda \circ g(j)$ for all $j \in J$. Now let $i \in I$ be arbitrary. We have that
\begin{multline*}
\phi(f + \lambda g)(e_i)
= (f + \lambda g)(\pi_1(\alpha(i)))(e_{\pi_2(\alpha(i))})\\
= (f(\pi_1(\alpha(i))) + \lambda \circ g(\pi_1(\alpha(i))))(e_{\pi_2(\alpha(i))})\\
= f(\pi_1(\alpha(i)))(e_{\pi_2(\alpha(i))}) + \lambda(g(\pi_1(\alpha(i))))(e_{\pi_2(\alpha(i))})\\
= \phi(f)(e_i) + \lambda(\phi(g)(e_i))\\
= \phi(f)(e_{\pi_2(\alpha(i))}) + (\lambda \circ \phi(g))(e_i)\\
= (\phi(f) + \lambda \circ \phi(g))(e_i)\\
= (\phi(f) + \lambda \phi(g))(e_i).
\end{multline*}
Since $i$ was arbitrary, we conclude that $\phi(f+\lambda g) = \phi(f) + \lambda \phi(g)$. Since $f$,$g$, and $\lambda$ were arbitrary, this shows that $\phi$ is an $R$-module homomorphism.
Edit: fixed some typos.
