I confronted a problem when I was reading the theorem stating that if $f_0$ and $f_1$ are homotopic maps (not to be confused with chain homotopy notion) from $X$ to $Y$, then they induce the same natural homomorphism from the homology group of $X$ to that of $Y$.
Using this theorem, I want to prove that the $n$th homology group of a convex subset $U$ of $\mathbb R^n$ is zero. In fact the identity map on $U$ and the map $c$ on $U$ whose image is a single point $x\in U$ are homotopic. Now if I prove that the map $c$ induces the zero homomorphism on the $n$th homology group of $U$ then we are done. How can I show that?
Best Answer
You can view the map $c$ as a composite: $U \to \{x\} \to U$. Then the induced map on homology looks like $H_n(U) \to H_n(\{x\}) \to H_n(U)$. If you can show that $H_n(\{x\}=0$, then the induced map on homology must be the zero homomorphism.