Why does the contractibility of $CX$ imply that $\pi_n(CX,X,x_0)\cong\pi_{n-1}(X,x_0)$?

Question

My Algebraic Topology notes say that because the cone of $X$ is contractible (which is easy enough to see, $H([x,t],s)=[x,t(1-s)]$ does the job) the Long Exact Sequence of Relative Homotopy Groups gives $\pi_n(CX,X,x_0)\cong\pi_{n-1}(X,x_0)$. I however fail to see how exactness gives us the required conclusion. What am I missing here? Thanks in advance.

Answer 1

Looking at a piece of the long exact sequence, we have:

$$ \pi_n(CX) \to \pi_n(CX,X) \to \pi_{n-1}(X) \to \pi_{n-1}(CX)$$

the first and last terms are zero by the contractibility of $CX$. This implies that the middle map is an injection and surjection of groups, and therefore an isomorphism.