Is this linear algebra proof correct? I did more linear algebra exercies and I'd appreciate if someone could tell me if my solution is correct. The exercise is: Show that if $T_1,...,T_n$ are injective linear maps (of the right dimensions) then $T_1 ...T_n$ is also injective.
My solution: Proceed by induction. The base case: Assume $T_1,T_2$ are injective. Consider $T_1T_2$. If this map is not injective then $T_1$  would not be injective. This would be a contradiction to the assumption. Therefore the product is injective. 
Assume $T_1....T_n$ is injective. Then $T_1....T_n$ and $T_{n+1}$ are injective linear maps. Applying the same argument as in the base case yields that $T_1...T_{n+1}$ is injective. Which is the claim. 
 A: If $T_1T_2$ is not injective, this does not imply that $T_1$ is not injective. Consider $T_1$ the identity map and $T_2 = 0$. 
The theorem you are trying to prove is not really one of linear algebra, and may be proved (using your approach!) for general maps on arbitrary sets $f_i:X_i\to X_{i+1}$. In the linear algebra case though, you might want to think about the rank of your homomorphisms and how that relates to composition, assuming we are talking about finite-dimensional vector spaces.
A: It's rather lacking on details in the base case. In general, non-injectivity of $T_1 T_2$ does not imply that $T_1$ is not injective - you have to use the fact that $T_2$ is injective to get this (example: $T_1 = $ identity, $T_2$ = zero map). It's also good practice not to use contradiction unless it's necessary.
As a better way, consider doing a bit of element following: Suppose that $T_1 T_2 (x) = 0$. Since $T_1$ is injective, $T_2(x) = 0$. Since $T_2(x) = 0$ and $T_2$ is injective $x = 0$. Hence $T_1 T_2$ is injective.
