I'm having significant trouble understanding how Hatcher determines that $ker(i_*)=(\alpha,-\alpha)$ (where the $i_*$ is the map appearing in the top row of the long exact sequence). I understand that $X\times \partial I$ is effectively the disjoint union of two copies of $X$ and so $H_n(X\times \partial I)\cong H_n(X)\oplus H_n(X)$ and Hatcher's proposed kernel makes sense if the map $i_*: H_n(X \times \partial I) \rightarrow H_n(X\times I)$ were defined by $(x,y) \mapsto x+y$, but I don't see why the induced map from the inclusion would be a sum.
What am I missing?
Best Answer
The map $H_n(X) \oplus H_n(X) \to H_n(X \times I)$ is $(i_0)_* \oplus (i_1)_*$, where $i_0$ and $i_1$ are the inclusions $X \to X \times I$ sending $x$ to $(x, 0)$ and $(x, 1)$ respectively. But $i_0$ and $i_1$ are homotopic (the identity map $X \times I \to X \times I$ is a homotopy), so the induced maps $(i_0)_*$ and $(i_1)_*$ are the same. So any $(\alpha, -\alpha) \in H_n(X) \oplus H_n(X)$ maps to $(i_0)_*(\alpha) - (i_1)_*(\alpha) = 0$.
Conversely, if $(\alpha, \beta)$ is in the kernel, we have $(i_0)_*(\alpha) + (i_1)_*(\beta) = 0$. However, if we apply $\pi_*$ to this, where $\pi : X \times I \to X$ is the projection, we get $$ 0 = (\pi i_0)_*(\alpha) + (\pi i_1)_*(\beta) = \alpha + \beta $$ since $\pi i_0$ and $\pi i_1$ are the identity map on $X$. Thus $\beta = -\alpha$, which is to say that anything in the kernel has the form $(\alpha, -\alpha)$ for some $\alpha \in H_n(X)$.