# Injective implies invertible? Injective and well-defined implies bijective?

I have two questions regarding functions regarding linear maps: (Let $X$ and $Y$ be to Banach spaces)

1. If $T:X\rightarrow Y$ is injective, then $T^{-1}$ exists, right?
2. If $T:X\rightarrow Y$ is injective and $T^{-1}$ well-defined, then $T$ is bijective, right?

Am I right in these questions? Or are there counter-examples? Thank you for your help!

• Forgot to mention that $T$ is bounded (making $T^{-1}$ bounded as well, right?) May 18 '15 at 6:42

No, if $T$ is for instance the inclusion of a proper subspace then $T$ is injective but not surjective. The inverse $T^{-1}$ cannot be well-defined unless $T$ is also surjective.
In other words, $T^{-1}$ is well-defined if and only if $T$ is injective and surjective. By the way, if $T$ is bounded then $T^{-1}$ is automatically bounded as well by the bounded inverse theorem.
If $T:X\to Y$ is injective, you can consider the application $T^*:X\to T(X)$. Then $T^*$ is bijective and has inverse..