Problem of the Week #113 - July 28th, 2014

  • MHB
  • Thread starter Chris L T521
  • Start date
In summary, the conversation was about an individual who is known for their exceptional summarizing skills, but lacks the ability to respond or reply to questions. They are only able to provide a summary of the content at hand.
  • #1
Chris L T521
Gold Member
MHB
915
0
Here's this week's problem!

-----

Problem
: Let $U$, $V$, and $W$ be three left $K$-vector spaces, and $\psi$, $\phi$ linear maps, fitting into a short exact sequence: $$0 \longrightarrow U \xrightarrow{~\psi~} V \xrightarrow{~\phi~} W \longrightarrow 0.$$
Define
$$S = \{\sigma \in \mathrm{Hom}_K(W,V) : \phi \circ \sigma = \text{Id}_W\}.$$
(An element of $S$ is called a splitting of the short exact sequence). Prove that there exists a bijection from $\mathrm{Hom}_K(W,U)$ to $S$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
  • #2
This week's problem was correctly answered by Euge and Opalg. You can find Opalg's solution below.

[sp]The key fact here is that if $Y$ is a $K$-vector subspace of $X$ then there exists a $K$-vector subspace $Z$ of $X$ such that $X$ is canonically isomorphic to the direct sum $Y\oplus Z$ (in category-theoretic terms, the category $\mathbf{Vect}_K$ admits coproducts). This is proved by taking a basis for $Y$, extending it to a basis for $X$ and defining $Z$ to be the subspace generated by all the additional vectors introduced into the basis. In general, when the spaces are not finite-dimensional, this will require the use of the axiom of choice.

Take $Y$ to be the subspace $\psi(U) \subseteq V$ and let $Z$ be a complementary subspace as above. Then $Y$ is isomorphic to $U$ (because $\psi$ is injective). Also, $Z$ is isomorphic to $W$ (because the restriction $\phi|_Z: Z\to W$ is surjective and its kernel is in $Y\cap Z = \{0\}$). Denote by $\rho: \phi(z) \mapsto z$ the inverse homomorphism from $W$ to $Z$.

Now suppose that $\theta \in \mathbf{Hom}_K(W,U)$. Then $\psi \circ \theta \in \mathbf{Hom}_K(W,V)$ and $\phi \circ (\psi \circ \theta) =(\phi \circ \psi) \circ \theta) = 0$ (by exactness at $Y$). Define $\sigma_\theta:W\to V$ by $\sigma_\theta(w) = (\psi\circ\theta)(w) + \rho(w)$. Then $(\phi\circ\sigma_\theta)(w) = (\phi \circ \psi \circ \theta)(w) + (\phi\circ \rho)(w) = 0+w=w.$ Therefore $\sigma_\theta \in S$. The map $\theta \mapsto \sigma_\theta$ is clearly injective because $\psi$ is injective.

Conversely, suppose that $\sigma \in S$. For $w\in W$, $\sigma(w) - \rho(w) \in \ker(\phi) = Y$. Define $\theta(w) = \psi^{-1}(\sigma(w) - \rho(w))$. Then $\theta \in \mathbf{Hom}_K(W,X)$ and $\sigma_\theta(w) = (\psi \circ \psi^{-1})(\sigma(w) - \rho(w)) + \rho(w) = \sigma(w) - \rho(w) + \rho(w) = \sigma(w)$. So $\sigma = \sigma_\theta$. Thus the map $\theta \mapsto \sigma_\theta$ is surjective, and consequently bijective, as required.[/sp]
 

FAQ: Problem of the Week #113 - July 28th, 2014

What is the "Problem of the Week #113 - July 28th, 2014"?

The "Problem of the Week #113 - July 28th, 2014" is a weekly challenge or puzzle posted by a scientific organization or community for individuals to solve and submit their solutions.

Who can participate in the "Problem of the Week #113 - July 28th, 2014"?

Anyone with an interest in science and problem-solving can participate in the "Problem of the Week #113 - July 28th, 2014". It is open to all ages and backgrounds.

How do I submit my solution for the "Problem of the Week #113 - July 28th, 2014"?

To submit your solution for the "Problem of the Week #113 - July 28th, 2014", you can follow the instructions provided by the organization or community hosting the challenge. This may include submitting your solution online, via email, or through a designated platform.

Are there any prizes for solving the "Problem of the Week #113 - July 28th, 2014"?

This may vary depending on the organization or community hosting the challenge. Some may offer prizes for the first person to solve the problem, while others may offer recognition or certificates of participation.

Can I collaborate with others to solve the "Problem of the Week #113 - July 28th, 2014"?

Again, this may vary depending on the rules set by the organization or community hosting the challenge. Some may allow collaboration, while others may require individual submissions. It is best to check the guidelines before collaborating with others.

Similar threads

Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Back
Top