Proof of Injectivity, Surjectivity and Isomorphism in Finite Dimensions

Routine facts about mutual inclusion of classes are used without comment (Toolkit for AL2). Let $T$ be a linear map from $V$ to $W$. Its kernel and image are subspaces (Kernel and Image of a Linear Map Are Subspaces (ii)), hence vector spaces with the operations of $V$ and of $W$ (Linear Subspaces Are Vector Spaces; Intersections (i)), and both are finite-dimensional (Properties of Dimension (v)).

Suppose that $T$ is injective. Then $\ker T=\{0\}$ (Kernel and Image of a Linear Map Are Subspaces (iii)), so $\dim\ker T=0$ (Properties of Dimension (iv), for the vector space $\ker T$). By The Rank–Nullity Theorem (i), $\dim V=\dim\operatorname{im}T$, and $\dim\operatorname{im}T\leq\dim W$ because $\operatorname{im}T$ is a subspace of $W$ (Properties of Dimension (v)). Hence $\dim V\leq\dim W$, which proves (i).

Suppose that $T$ is surjective. Then $\operatorname{im}T=W$ (Kernel and Image of a Linear Map Are Subspaces (iv)), and The Rank–Nullity Theorem (i) gives

$$\dim V=\dim\ker T+\dim W\geq\dim W,$$

which proves (ii).

Now let $\dim V=\dim W$. By The Rank–Nullity Theorem (i), $\dim W=\dim\ker T+\dim\operatorname{im}T$, so $\dim\ker T=0$ if and only if $\dim\operatorname{im}T=\dim W$. The first equality says that $\ker T=\{0\}$ (Properties of Dimension (iv)), that is, that $T$ is injective (Kernel and Image of a Linear Map Are Subspaces (iii)); the second says that $\operatorname{im}T=W$ (the equality case of Properties of Dimension (v)), that is, that $T$ is surjective (Kernel and Image of a Linear Map Are Subspaces (iv)). Since bijective means injective and surjective (Injective, Surjective and Bijective Functions between Classes (iii)), the three properties are equivalent, which proves (iii).

If $V$ and $W$ are isomorphic, there is an isomorphism of $V$ onto $W$, that is, a bijective linear map (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (iii)), and (i) and (ii) give $\dim V=\dim W$. Conversely, let $\dim V=\dim W=n$. Choose bases $(v_{1},\dots,v_{n})$ of $V$ and $(w_{1},\dots,w_{n})$ of $W$ (Existence and Invariance of the Length of a Basis (i), Finite-Dimensional Vector Spaces and Their Dimension (ii)), and let $T$ be the linear map with $Tv_{i}=w_{i}$ for every $i$ (Linear Maps Are Determined by Their Values on a Basis (i)). Its image is a subspace of $W$ (Kernel and Image of a Linear Map Are Subspaces (ii)) containing every $w_{i}=Tv_{i}$, hence containing their span (The Span of a Finite Family is the Smallest Subspace Containing It (iii)), which is $W$ (Finite Basis of a Vector Space (i)). So $T$ is surjective, hence bijective by (iii), hence an isomorphism, and $V$ and $W$ are isomorphic. This proves (iv).

Finally, let $T\in\mathcal{L}(V)$. By Linear Maps Form a Vector Space and a Monoid under Composition (vi), $T$ is a unit of $\mathcal{L}(V)$ if and only if it is bijective. By (iii) with $W=V$, $T$ is bijective if and only if it is injective, and if and only if it is surjective; and these say that $\ker T=\{0\}$ (Kernel and Image of a Linear Map Are Subspaces (iii)) and that $\operatorname{im}T=V$ (Kernel and Image of a Linear Map Are Subspaces (iv)). This proves (v).