[proofplan] Part (i) applies $\alpha$ to a dependence relation and uses injectivity ($\ker\alpha = \{\mathbf{0}\}$) to force all coefficients to vanish. Part (ii) uses surjectivity to lift any vector in $V$ to a preimage and then expresses it via the spanning set. Part (iii) combines (i) and (ii) using the fact that isomorphisms are bijective. [/proofplan] [step:Show injective linear maps preserve linear independence (part (i))] Let $S \subset U$ be linearly independent and let $\alpha: U \to V$ be injective. Consider distinct elements $s_1, \ldots, s_n \in S$ (the elements $\alpha(s_1), \ldots, \alpha(s_n)$ are distinct in $\alpha(S)$ since $\alpha$ is injective). Suppose \begin{align*} \sum_{i=1}^{n}\lambda_i \alpha(s_i) = \mathbf{0}_V. \end{align*} By linearity of $\alpha$: \begin{align*} \alpha\left(\sum_{i=1}^{n}\lambda_i s_i\right) = \mathbf{0}_V. \end{align*} Since $\alpha$ is injective and $\alpha(\mathbf{0}_U) = \mathbf{0}_V$, we have $\sum_{i=1}^{n}\lambda_i s_i = \mathbf{0}_U$. By linear independence of $S$, $\lambda_i = 0$ for all $i$. Hence $\alpha(S)$ is linearly independent. [/step] [step:Show surjective linear maps preserve spanning (part (ii))] Let $S \subset U$ span $U$ and let $\alpha: U \to V$ be surjective. Let $v \in V$. By surjectivity, there exists $u \in U$ with $\alpha(u) = v$. Since $S$ spans $U$, there exist $s_1, \ldots, s_n \in S$ and $\lambda_1, \ldots, \lambda_n \in \mathbb{F}$ with $u = \sum_{i=1}^{n}\lambda_i s_i$. By linearity: \begin{align*} v = \alpha(u) = \alpha\left(\sum_{i=1}^{n}\lambda_i s_i\right) = \sum_{i=1}^{n}\lambda_i \alpha(s_i). \end{align*} Hence $v \in \langle\alpha(S)\rangle$, and since $v$ was arbitrary, $\alpha(S)$ spans $V$. [/step] [step:Combine to show isomorphisms preserve bases (part (iii))] If $\alpha: U \to V$ is an isomorphism, then $\alpha$ is bijective by the [Isomorphism iff Linear Bijection](/theorems/378) theorem, hence both injective and surjective. If $S$ is a basis for $U$, then $S$ is linearly independent and spans $U$. By part (i), $\alpha(S)$ is linearly independent in $V$. By part (ii), $\alpha(S)$ spans $V$. Hence $\alpha(S)$ is a basis for $V$. [/step]