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Let $h:S^1\to X$ a continuous map. Show that if $h$ can be extended to a continuous map $H:B^2\to X$, then $h_*$ is a trivial homomorphism.

I simply don't know how to use the fact that h can be extended to a continuous map H to prove that $h_*$ is a trivial homomorphism. We have to prove that $h_*([f])=c_{x_0}$, where $c_{x_0}$ is constant map from $I$ to the base point $x_0\in X$ and $f$ is a loop in $S^1$. I tried to use the definition of $h_*([f])=[h\circ f]$ also no results. I'm think about to use deformations retractions of $D^2$ to a point but I don't know how.

I need a hand here


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Hint: Homotopic maps induce the same map on fundamental group. – Jason DeVito Nov 12 '12 at 3:02
Second hint: the question you asked earlier:… – user27126 Nov 12 '12 at 3:04
@Sanchez are you speaking about the fact $D^2$ is contractible? yes I know I have to use this, but I don't know how to use in this question. – user42912 Nov 12 '12 at 3:19
@JasonDeVito can you explain more? or at least say where I can find this, I'm studying the Lee's book and I've never seen this. – user42912 Nov 12 '12 at 3:22
@user42912: Hatcher's Algebraic Topology book (freely available on his website) has the proof. But for now, ignore it: Chris's answer is much better than my hint. – Jason DeVito Nov 12 '12 at 3:56
up vote 3 down vote accepted

Compute the induced map on $\pi_1$, noting that $h$ factors through $H$ by definition and that $\pi_1$ respects composition.

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sorry, can you please explain a little bit more? I didn't understand what you said. – user42912 Nov 12 '12 at 4:01
I can, but I need to know which part and precisely what your confusion is. – Chris Gerig Nov 12 '12 at 4:19
what do you mean by induced map on $\pi_1$, which map? and this sentence $"h factors through H"? almost everything, sorry. – user42912 Nov 12 '12 at 4:26
I highly advise you spend the time learning this terminology, as it is required to even begin speaking of this subject. Given a map $f:A\to B$, the "induced map on" $\mathcal{H}$ (some operator or tool) is $\mathcal{H}(f):\mathcal{H}(A)\to\mathcal{H}(B)$, when appropriately defined. A map $f:A\to C$ "factors through" a map $g:B\to C$ when we can rewrite $f$ as the composition $A\to B\to C$ where the second arrow is $g$. – Chris Gerig Nov 12 '12 at 4:36
Thank you for the terminology. So when you say "Compute the induced map on $\pi_1$" you're saying to take $\pi_1(S^1)\to \pi_1(X)$? I know this, but how can I use the hypothesis with this? it's my problem. – user42912 Nov 12 '12 at 4:55

Let $f$ be a loop at $x_0$, viz. $f(0) = f(1) = x_0$. Now for all practical purposes we may identify the points $0$ and $1$ on $\partial I$ and say now that $f$ is a map from $S^1$ to $X$ such that $f(z_0) = x_0$. We call $z_0$ the single point obtained from gluing together the endpoints of the interval $[0,1]$. Now to prove that $f$ is homotopic to the constant map $c$ that sends every point of $S^1$ to $x_0$, what we really mean is that

$$f \simeq c \hspace{3mm} \textrm{rel} \{z_0\}.$$

By assumption there is a map $g :D^2 \to X$ that extends $f$. Geometrically, what we will do now is contract the disk to the point $z_0$ using straight line homotopy. This is the key point, taking advantage of the fact that $D^2$ is a convex set. We now turn our attention to proving this relative homotopy: Define

$$F(x,t) : D^2 \times I \to X$$

\noindent by $F(x,t) = g((1-t)(x-z_0) + z_0) = g(x(1-t) + tz_0)$. It is clear that $F$ is a continuous function. Then because $D^2$ is convex the straight line $x(1-t) + tz_0$ connecting any $x \in D^2$ to $z_0$ on the boundary is still in $D^2$ so it makes sense to apply $g$ to such an expression. It is clear that $F(x,0) = g(x)$ which when restricted to $S^1$ is $f(x)$. Furthermore $F(x,1) = g(z_0) = x_0$. It now remains to check for all $t \in [0,1]$ that $F(z_0,t) = x_0$. But this is clear because $F(z_0,t) = g((1-t)(0) + z_0) = g(z_0) = x_0$ showing that $F|_{S^1 \times I}$ gives a homotopy between $f$ and $c$ relative to the subspace $\{z_0\}$ of $S^1$.

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