Show that $Im T$ and $U/Nuc T$ are isomorphic for a linear transformation $T: U \longrightarrow V$

Question

Show that $Im$ $T$ and $U/Nuc$ $T$ are isomorphic for a linear transformation $T: U \longrightarrow V$

Hi guys, I know how to show this for vectorial spaces with finite dimension, but I don't have idea how to show this when vectorial spaces have infinite dimension. Someone can give me a hint, please?

Thanks!

My attempt for vectorial spaces with finite dimension:

If $U$ have finite dimension, so $dim$ $U$ $=$ $dim$ $Nuc$ $T$ $+$ $dim$ $U/Nuc$ $T$ and $dim$ $U$ $=$ $dim$ $Nuc$ $T$ $+$ $dim$ $Im$ $T$, therefore $dim$ $Im$ $T$ $=$ $dim$ $U/Nuc$ $T$

If $U$ have finite dimension, $Im$ $T$ and $U/Nuc$ $T$ have finite dimension, so $Im$ $T$ and $U/Nuc$ $T$ are isomorphic.

Answer 1

The idea is to define $\overline{T}\colon U/\ker T \to V$ by $\overline{T}([x]) = T(x)$. You have to check that

• $\overline{T}$ is well-defined, that is, $[x] = [y] \implies T(x) = T(y)$;

• $\overline{T}$ is injective (take $[x] \in \ker \overline{T}$ and check that $[x] = [0]$, in other words, that $x\in \ker T$, which is almost trivial);

• ${\rm Im}(\overline{T}) = {\rm Im}(T)$

Note that the dimensions of the spaces involved are not relevant for this approach.