Do module isomorphisms preserve freeness?

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Let $f$ be module isomorphism $f:M_1 \rightarrow M_2$, with $M_1$ free. Then, is $M_2$ free too?
I tried to come up with a proof: Let $B_1$ be the basis of $M_1$. Then we know that $f(B_1)$ is linearly independent. $$f(m_1)=f(\sum r_ib_{1i})=\sum r_if(b_{1i})$$ for any $m_1 \in M_1$ and $b_{1i} \in B_1$. Considering the fact that $f$ is bijective, is this enough to prove my question?

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The proof is correct. In general all properties of objects that make sense must be preserved by isomorphisms. Moreover, in categories that are not small there is a problem with defining equality of objects, since it must by definition the equality is a relation on the class of objects. Since a relation is a subset of the product, it is not well defined when objects form a proper class, in particular in the case of modules.