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
 A: Compute the induced map on $\pi_1$, noting that $h$ factors through $H$ by definition and that $\pi_1$ respects composition.
A: 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 $S^1$. 
