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I have two problems that I don't know how to do )=.

i) Let X path connected. Let $x_0 , x_1 \in X$. Show that $\pi _1 \left( {X,x_0 } \right)$ is abelian iff for every $\alpha, \beta$ paths from $x_0$ to $x_1$ , we have $\widehat\alpha = \widehat\beta$.

Where the hat denotes the homomorphism: $$ \eqalign{ & \widehat\alpha :\pi _1 \left( {X,x_0 } \right) \to \pi _1 \left( {X,x_1 } \right) \cr & \widehat\alpha \left( {\left[ f \right]} \right) = \left[ \alpha \right]^{ - 1} \left[ f \right]\left[ \alpha \right] \cr} $$

I did the side where the group is abelian, but the other I could not )=.

ii) The other is a property that I read in wikipedia. If $X$ and $Y$ are path connected, then $$ \pi _1 \left( {X\times Y} \right) \cong \pi _1 \left( {X\,} \right) \times \pi _1 \left( Y \right) $$ I'm not sure if this nice property is easy to show. )=

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(ii) is a proposition in Hatcher's Algebraic Topology. – HbCwiRoJDp Feb 23 '12 at 23:28

For $i$) "the other", I assume you're asking to show that if $\widehat{\alpha} = \widehat{\beta}$ for all $\alpha$ and $\beta$, then $\pi_1(X,x_0)$ is abelian.

Here's a hint: Let $a,b\in \pi_1(X,x_0)$. You want to show that $aba^{-1} = b$. Is there a path $\alpha$ starting and ending at $x_0$ for which $\widehat{\alpha}(b) = aba^{-1}$? Is there another easy path $\alpha'$ from $x_0$ to $x_0$ for which $\widehat{\alpha'}b = b$?

For $ii$), the point is that a map from $S^1$ into $X\times Y$ is nothing more than 2 maps, one from $S^1$ into $X$ and the other from $S^1$ into $Y$.

As a hint: If $\pi$ and $\pi'$ are the projections from $X\times Y$ onto $X$ and $Y$, respectively, then show $[\gamma]\rightarrow(\pi_*[\gamma],\pi'_*[\gamma])$ is an isomorphism from $\pi_1(X\times Y)$ to $\pi_1(X)\times\pi_1(Y)$.

Edit It seems there is a bit of confusion as to whether or not the correct assumption is "For all $x_0$ and $x_1$, for any two paths between them the induced maps are the same" or if it's "There exists two points $x_0$ and $x_1$ so that any 2 paths between them induce the same map." I claim that the two assumptions are actually equivalent.

More specifically, in a path connected space $X$, if there is some pair of points $x_0$ and $x_1$ so that all paths between $x_0$ and $x_1$ induce the same map on $\pi_1$, then for all pairs of points $x$ and $y$, any path between them induces the same map on $\pi_1$.

Once and for all, pick a path $\gamma_1$ from $x_0$ to $x$ and a path $\gamma_2$ from $y$ to $x_1$.

Let $\alpha_1$ and $\alpha_2$ be two paths from $x$ to $y$. Then the path $\alpha_I' = \gamma_1 * \alpha_i * \gamma_2$ is a path from $x_0$ to $x_1$.

Finally, let $f\in\pi_1(X,x)$. Then $[\gamma_1 f \gamma_1^{-1}]\in \pi_1(X,x_0)$. By assumption, $\widehat{\alpha_1'}[\gamma_1 f \gamma_1^{-1}] = \widehat{\alpha_2'}[\gamma_1 f \gamma_1^{-1}]$.

Cruching these two things and simplifying gives $$[\gamma_2^{-1} \alpha_1^{-1} f \alpha_1 \gamma_2] = [\gamma_2^{-1} \alpha_2^{-1} f \alpha_2 \gamma_2].$$ Applying $\widehat{\gamma_2^{-1}}$ to both sides shows that $\widehat{\alpha_1} = \widehat{\alpha_2}$ as claimed.

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Shouldn't the paths $\alpha$ and $\alpha '$ be from $x_0$ to $x_1$ in the first question? (sorry to resurrect this question, I was referred to this post from a post that I posted today. – MathematicalPhysicist Nov 9 '11 at 8:34
@MathematicalPHysicist: I'm taking the assumption as "for all $x_0$ and $x_1$ blah blah" so I'm free to choose $x_0 = x_1$. But you're right that it could mean "for a fixed $x_0$ and $x_1$ blah blah". I'll think about that case and add an edit as soon as I can. – Jason DeVito Nov 9 '11 at 14:32
OK, I guess your'e right and they meant for all, and not some fixed points, cause in Hatcher's book he talks of $\pi_1(X)$ without any mention of the points. Thanks for clearing it out. – MathematicalPhysicist Nov 9 '11 at 14:37
@MathematicalPhysicist: It's updated. Let me know if it's still unclear. – Jason DeVito Nov 9 '11 at 14:55

Following what Jason says, maybe you could start proving that you already have a bijection between sets of continuous maps

$$ \Phi : \mathrm{Hom}(Z, X\times Y) \longrightarrow \mathrm{Hom} (Z, X) \times \mathrm{Hom}\mathrm(Z,Y) $$

for any topological space $Z$, particularly for $Z = S^1$.

This is just the universal property of the product topology: a continuous map $f : Z \longrightarrow X \times Y$ is the same as a pair of maps $Z \longrightarrow X$ and $Z \longrightarrow Y$. Namely, $\Phi$ sends $f$ to $(p_Xf, p_Yf)$, where $p_X : X\times Y \longrightarrow X$ and $p_Y : X\times Y \longrightarrow Y$ are the natural projections. You should think who is the inverse $\Psi$ of this $\Phi$.

The next step then could be to show that (punctured) homotopy relationship preserves this bijection.

Namely, you should prove that if you have an homotopy $H: Z\times I \longrightarrow X\times Y$ between maps $f, g: Z \longrightarrow X\times Y$, then you also have homotopies between compositions: $p_Xf \sim p_X g$ and $p_Yf \sim p_Yg$.

Adding preserving base point maps and preserving base point homotopies at this stage should be harmless.

As a consequence, you will have a well-defined map between homotopy classes induced by $\Phi$:

$$ \widetilde{\Phi} : [(Z,z_0), (X\times Y,(x_0,y_0))] \longrightarrow [(Z,z_0), (X,x_0)] \times [(Z,z_0), (Y,y_0)] \ . $$


$$ \widetilde{\Phi}([f]) = ([p_Xf], [p_Yf]) \ . $$

(Here $x_0 \in X$, $y_0 \in Y$ and $z_0 \in Z$ are the base points.)

Finally, in order to get an inverse for this map between sets of (punctured) homotopy classes, you could repeat the same process for $\Psi$, obtaining some $\widetilde{\Psi}$ in the opposite direction.

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Where you used the fact , that the spaces are path connected? – Daniel Nov 1 '11 at 22:08
We need not the spaces to be path connected here. – a.r. Nov 2 '11 at 2:29

The more general result on products is that if $\pi_1(X)$ denotes the fundamental groupoid of the space $X$, then for spaces $X,Y$, the natural morphism

$$\pi_1(X \times Y) \to \pi_1(X) \times \pi_1(Y) $$ determined by the projections $X \times Y \to X, X \times Y \to Y$, is an isomorphism. ("Topology and groupoids" 6.4.4).

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