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I'm having troubles in understanding a rapidly solved exercise, which is actually a sort of chain of consequences. It says: be $A; B$ two matrices $n\times n$ over a field $K$ and be $F_A, F_B$ their associated linear transformations. Then:

  • $AB$ is invertible, then $A$ and $B$ are invertible. $\color{blue}{\text{This is ok}}$.
  • $AB$ is invertible iff $F_{AB}: K^n \to K^n$ is an isomorphism. $\color{blue}{\text{This is ok}}$ (where $F_{AB} = F_A \circ F_B$).
  • Having $F_{AB}$ then $F_B$ is injective and $F_A$ is surjective. $\color{red}{\text{This is not ok}}$.

Here I have problems in understanding. What does ensure me that if $F_{AB}$ is invertible, then $F_B$ is injective and $F_A$ is surjective? Why can't I have other combinations like the opposite one?

The only possible explanation I managed to give is this: in order for $F_{AB}$ to exist, we need something like $F_A: K^n \to K^p$ and $F_B: K^q \to K^n$, in this way $F_{AB}: K^p \to K^n$. This actually requires (for our purposes) $p = q = n$ $\color{red}{\text{Is this right?}}$.

In this way if we work on $K^n$ then $F_B$ is surely surjective because the codomain and the image are the same. I don't understand why $F_A$ is injective though. I'm losing myself in a glass of water perhaps...

Gerr
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    Why is it not OK? If $f\circ g=id$, then always $g$ is injective, see here, and $f$ is surjective. This is precalculus. With $f=F_A$, $g=F_B$. – Dietrich Burde Dec 27 '24 at 15:36
  • Just to clarify, there are two important facts. 1. $A$ is invertible iff $F_A$ is isomorphic. 2. The composition $F_{A}\circ F_{B}$ is $F_{AB}$ where $F_{AB}$ is transformation of $AB$. – Yos Dec 27 '24 at 15:46
  • @DietrichBurde Oh, you're right. I lost myself in nothing. I got it now! Thanks! Case closed! – Gerr Dec 27 '24 at 15:47

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What you are saying hold more in general, not just for linear transformations.

Let $f: Y \rightarrow Z$ and $g: X \rightarrow Y$ be two functions, and assume that $f \circ g$ is invertible (equivalently, it is an isomorphism). Then $g$ is injective and $f$ is surjective.

The proof is quite easy. Here for the sake of completeness.

To show that $g$ is injective, we need to prove that if $g(x_1) = g(x_2)$ for some $x_1, x_2 \in X$, then $x_1 = x_2$. Assume $g(x_1) = g(x_2)$. Applying $f$ to both sides, we have $f(g(x_1)) = f(g(x_2))$.
Since $f \circ g$ is invertible, this implies $x_1 = x_2$. Therefore, $g$ is injective.

To show that $f$ is surjective, we need to prove that for every $z \in Z$, there exists $y \in Y$ such that $f(y) = z$. Since $f \circ g$ is invertible, for every $z \in Z$, there exists $x \in X$ such that $(f \circ g)(x) = z$.
Explicitly, $f(g(x)) = z$. Setting $y = g(x)$, we have $f(y) = z$.
Thus, $f$ is surjective.

An analogue result holds also in the context of category theory.