Proof - Inverse of linear function is linear

Question

This is my first proof related to linear functions. It refers to the linear-algebra-$\textit{linear}$ (not the calculus-$\textit{linear}$). Please comment.

Theorem

The inverse of a linear bijection is linear.

Proof

Let $X,Y$ be vector spaces over a common field. Let $f : X \rightarrow Y$ be a linear bijection. We denote by $f^{-1}$ the inverse of $f$. It remains to prove that $f^{-1}$ is linear, i.e. both $\textit{additive}$ and $\textit{homogeneous}$.

Additivity

Let $y_1, y_2 \in Y$. We prove that $$f^{-1}(y_1 + y_2) = f^{-1}(y_1) + f^{-1}(y_2).$$ \begin{equation*} \begin{split} f^{-1}(y_1) + f^{-1}(y_2) &= f^{-1}\Big( f\big( f^{-1}(y_1) + f^{-1}(y_2) \big) \Big) && \quad \text{by bijectivity} \\ &= f^{-1}\Big( f\big( f^{-1}(y_1) \big) + f\big( f^{-1}(y_2) \big) \Big) && \quad \text{by linearity of } f \\ &= f^{-1}\Big( y_1 + f\big( f^{-1}(y_2) \big) \Big) && \quad \text{by bijectivity} \\ &= f^{-1}(y_1 + y_2) && \quad \text{by bijectivity}\phantom{\Big(\Big)} \\ \end{split} \end{equation*}

Homogeneity

Let $y \in Y$ and let $s$ be a scalar. We prove that $$f^{-1}(sy) = sf^{-1}(y).$$ \begin{equation*} \begin{split} sf^{-1}(y) &= f^{-1}\Big( f\big( sf^{-1}(y) \big) \Big) && \quad \text{by bijectivity} \\ &= f^{-1}\Big( sf\big( f^{-1}(y) \big) \Big) && \quad \text{by linearity of } f \\ &= f^{-1}(sy) && \quad \text{by bijectivity}\phantom{\Big(\Big)} \\ \end{split} \end{equation*}

QED

Answer 1

By definition a linear map holds the following:

$i) F(a + b)= F(a) + F(b) \\ ii)F(ka)=kF(a)$ where $a \in V$, V is a vector space and $k \in K$, K is a field.

The theorem above shows that the function $f^{-1}$ preserves both vector addition and scalar multiplication.

Answer 2

General principle: You can demonstrate two vectors in X are equal by showing that f maps them to the same vector in Y (since f is injective). For example: Apply $f$ to $f^{-1}(y_1)+f^{-1}(y_2)$ and argue that the result is $y_1 + y_2$. Do a similar calculation involving $f^{-1}(y_1+y_2)$. You should then be able to see why the inverse of an additive injection is additive. Homogeneity can be approached the same way. You may observe that surjectivity of $f$ is not really involved in what is going on here.