# A question on Differential Geometry

I would thank so much any help with this problem. I have thinking a lot and have no idea what to do.

Suppose that $V$ and $W$ are vector spaces (not necessarily finite dimensional) and $\Phi \colon V \to W$ is a suryective linear map.

I have to find:

1. A vector space $S$.

2. A linear map $\Psi \colon S \to W$.

3. An isomorphism $\Theta \colon S\times W \to V$.

Such that $\Theta(\Gamma_\Psi) = \ker(\Phi)$, where $\Gamma_\Psi \subset S\times W$ is the graph of $\Psi$.

I don't know what to do or how to proceed with the problem (not even what $S$ to use).

Some help will be so appreciated.

Thanks.

• Why are you tagging this with differential geometry? Sep 19, 2017 at 1:31
• I know it is not about that area but it this is a problem that I have to do for this course... ¿Should I remove that tag? Sep 19, 2017 at 1:35
• Sorry, This kind of problems are weird for me and didn't know what it is really about. Sep 19, 2017 at 1:37

Let $S$ be the kernel of $\Phi$, and let $\Psi$ be the trivial map. The short exact sequence of vector spaces $$0 \to S \to V \xrightarrow{\Phi} W \to 0$$ splits like any other, so there is an isomorphism $\Theta:S \times W \to V$ that fixes $S \times \{ 0 \} \cong S$. The graph of the trivial map $\Psi$ is just $S \times \{ 0 \}$, so we are done.
• I can't believe it. Thanks, I must see that $S$ was (or is) the Kernel... Sep 19, 2017 at 2:01
• @JPLK The short answer is that vector spaces are free modules, hence projective modules, and any short exact sequence terminating in a projective module splits. In more detail, we may explicitly define a retract $V \to S$ of the inclusion $S \to V$ by extending a basis of $S$ to all of $V$ and letting the retract map all the basis vectors that aren't in $S$ to $0$. Or as another alternative, we may define a section $W \to V$ of $\Phi$ by picking preimage vectors of a chosen basis of $W$. Sep 19, 2017 at 21:00