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!

  • $\begingroup$ 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

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.


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