2
$\begingroup$

I apologize in advance I haven't my own approach here, it doesn't mean I haven't tried at all, I've just realized my lack of knowledge and practice in the subject. It would be great if I can get an answer t the following so I can closely study it. Meanwhile, I'll keep trying and edit for updates if any as I go on my own.

a. For any spaces $X,Y$, with basepoints $x \in X$ and $y \in Y$, construct inverse bijections

$$\theta:\pi_1(X \times Y, (x,y)) \rightarrow \pi_1(X,x) \times \pi_1(Y,y)$$ $$\phi:\pi_1(X,x) \times \pi_1(Y,y) \rightarrow \pi_1(X \times Y,(x,y))$$

which are isomorphisms of groups. You do not need to show that the group laws are defined but define $\phi,\theta$ explicitly.

b. Compute the fundamental group $\pi_1(A,a)$ of

$$A=\{(x,y) \in \mathbb{R}^2| y>1 \text{ and } x>0\}$$

where $a=(1,2) \in A$

For $a$, I was thinking(guessing rather) by drawing pictures. But to me, the "isomorphism" bit makes it harder...

Does anyone have a method to solve this? Help is very much appreciated!

$\endgroup$
  • $\begingroup$ I'm not sure it's clear what you are asking. Please make that a bit more explicit. $\endgroup$ – parsiad Feb 24 '16 at 15:04
  • $\begingroup$ If I deduce correctly in (b) it is the trivial group as $\;A\;$ is simply connected.. $\endgroup$ – DonAntonio Feb 24 '16 at 15:41
  • $\begingroup$ For a, you can apply the projections to the two factor spaces to any curve $\gamma \in\pi _1$. The two "obvious" maps are isomorphims because they are inverses of each other. - For b you might use a. $\endgroup$ – Hagen von Eitzen Feb 24 '16 at 16:57
  • $\begingroup$ Elements in $\pi_1(Z,z)$ are represented by loops at $z$, that is continuous maps $f:[0,1]\to Z$ with $f(0)=f(1)=z$. So, given such a map $f$ to $(X\times Y, (x,y))$, by projection, you get maps to $(X,x)$ and $(Y,y)$. Similarly you can define the inverse. $\endgroup$ – Mohan Feb 24 '16 at 17:00
1
$\begingroup$

First of all here is an hint to solve the problem a).

As a matter of notation in what follows $\mathbf{Top}[A,B]$ is the set of continuous functions from the topological space $A$ to the space $B$ and $I$ is the unit interval, a.k.a. $[0,1]$.

The product space $X \times Y$ comes equipped with two canonical projections $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$.

These projections induce two mappings $${\pi_X}_* \colon \mathbf{Top}[I,X \times Y] \to \mathbf{Top}[I,X]$$ and $${\pi_Y}_* \colon \mathbf{Top}[I,X \times Y] \to \mathbf {Top}[I,Y]$$, which are defined as ${\pi_X}_*(\alpha)=\pi_X \circ \alpha$ and ${\pi_Y}_*(\alpha)=\pi_Y \colon \alpha$.

This gives you a map $$\pi_* \colon \mathbf{Top}[I, X \times Y] \longrightarrow \mathbf{Top}[I,X] \times \mathbf{Top}[I,Y]$$ defined by $\pi_*(\alpha)=(\pi_X\circ\alpha,\pi_Y\circ\alpha)$.

It is not hard to prove that $\pi_*$ is a bijection (for instance using the universal property of the projections) and that it and its inverse restrict to two bijectives mappings between the sets of loops $\Omega(X\times Y,(x,y))$ and $\Omega(X,x)\times\Omega(Y,y)$.

Using the definition of $\pi_1(A,a)$ as a quotient of a subset $\Omega(A,a) \subseteq \mathbf{Top}[I,A]$, the space of loops at $a$, you should be able to prove that $\pi_*$ and $\pi_*^{-1}$ induce the isomorphisms $\theta$ and $\phi$.

About question b) the answer depends on what you have seen on algebraic topology till now.

Hope this helps and feel free to ask for additional details or clarifications.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.