# induced homomorphisms and extended maps

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

Thanks

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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: math.stackexchange.com/questions/235095/… – 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

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