Proof that composition of invertible linear transformations is invertible (without determinants) A crucial concept in linear algebra is that the composition of two invertible linear transformations is itself invertible. Here is the first proof I learned of this fact:
Proof: Suppose that $T_1: \mathbb{C}^n \to \mathbb{C}^n$ and $T_2: \mathbb{C}^n \to \mathbb{C}^n$ are both invertible with respective matrices $A_1$ and $A_2$. Then the matrix of their composition $T_2 \circ T_1$ is simply $A_2A_1$. Since $T_1$ and $T_2$ are invertible, we know that $\det(A_1) \neq 0$ and $\det(A_2) \neq 0$. Thus, we see that $\det(A_1A_2) = \det(A_1)\det(A_2) \neq 0$. Thus, the composition is also invertible. $\square$
I'm trying to now give a proof of this fact without using determinants. Any idea on where to start?
 A: In finite dimensional space it suffices to prove that $T_2\circ T_1$ is injective:
$$ x\in\ker(T_2\circ T_1)\iff T_2(T_1(x))=0\iff T_1(x)=0\iff x=0 $$
We used in the second equivalence the fact that $T_2$ is injective and in the last equivalence the fact that $T_1$ is injective. Conclude.
A: This much is true for any two invertible (i.e., bijective) functions, since $\;f\circ g\;$  is bijective if both $\;f\,,\,\,g\;$ are, and we get, using associativity of functions composition, that:
$$(f\circ g)\circ(g^{-1}\circ f^{-1})=f\circ\left(g\circ g^{-1}\right)\circ f^{-1}=f\circ \text{Id}\circ f^{-1}=f\circ f^{-1}=\text{Id}$$
and thus $\;f\circ g\;$ is invertible, and from the above it follows, by uniqueness of inverse of bijective functions, that in fact $\;g^{-1}\circ f^{-1}=(f\circ g)^{-1}\;$ 
A: If $T_1,T_2$ are invertible then $T_1^{-1},T_2^{-1}$ exist.
Hence $,T_2^{-1}T_1^{-1}$ exists and now only check that
$$(T_1T_2)(T_2^{-1}T_1^{-1})=Id.$$
A: If $A_1$ and $A_2$ are invertible, then they have inverses $A_1^{-1}$ and $A_2^{-1}$.  Then
$$
(A_2A_1)(A_1^{-1}A_2^{-1})=A_2(A_1A_1^{-1})A_2^{-1}=A_2IA_2^{-1}=A_2A_2^{-1}=I.
$$
Thus the composition has an inverse.
