Invertible and bijective on the context of linear algebra and general one.

Question

I'm reading linear algebra, and trying to figure out the relation between bijective(one-to-one and onto) and invertible. I've searched about similar questions and answers of them but I found something differ:

In the first comment of https://math.stackexchange.com/q/530446/390226, I read:

$$\textrm{["]invertible $\iff$ bijective[."].}$$

But I found another answer, https://math.stackexchange.com/a/2415538/390226, I read:

$$\textrm{["]invertible $\Longleftarrow$ bijective[."].}$$

I think they're both right since the first one is about linear transformation and the second one is about a general function. But I don't know the reason/details the two differ.

Answer 1

Bijective implies that there is an inverse function. Define $f^{-1}(x)$ to be the only solution of $f(y)=x$. Such solution exists for each $x$, due to the $f$ being onto, and the solution is unique due to $f$ being one-to-one.

In the context of linear algebra one wants to restrict attention to functions that are linear. A linear and bijective function would have a linear-algebra inverse, by definition, if it has an inverse function that is also linear.

Theorem: A linear function that has an inverse function, has a linear inverse function.

Proof: We just need to show that the inverse function $f^{-1}$ is linear. For $x,y$ vectors and $a$ scalar we have that there are some $X,Y$ such that $f(X)=x$ and $f(Y)=y$. Then $$f^{-1}(ax+y)=f^{-1}(af(X)+f(Y))=f^{-1}(f(aX+Y))=aX+Y=af^{-1}(x)+f^{-1}(y)$$

Therefore, $f^{-1}$ is linear.

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Linear algebra is special in this sense. In some other contexts the additional property is not necessarily shared by the inverse function. For example, in topology one cares about continuous functions. A continuous function can have an inverse, but that inverse not be continuous, like $f:[0,2\pi)\to \{(x,y)\in\mathbb{R}^2:\ x^2+y^2=1\}$ defined by $f(x)=(\cos(x),\sin(x))$.