Relative homology Let $E$ be a real banach space 
if $E=Y\oplus Z$ and if $S^{m-1}$ is the sphere on $Y$ ($\dim Y =m $) 
why  $H_{m-1}(E \setminus Z)\simeq H_{m-1}(S^{m-1})$ ?
please thank you
 A: A possible solution is to notice that $$h : \left\{ \begin{array}{ccc} E \backslash Z & \to & S^{m-1} \\ (y,z) & \mapsto & \left( \frac{y}{\| y\|},0 \right) \end{array} \right.$$ is a deformation retract of $E \backslash Z$ on $S^{m-1}$. For that, let $$f : \left\{ \begin{array}{ccc} S^{m-1} & \to & E \backslash Z \\ (y,0)& \mapsto & (y,0) \end{array} \right. \ \text{and} \ H : \left\{ \begin{array}{ccc} [0,1] \times E \backslash Z & \to & E \backslash Z \\ (t,y,z) & \mapsto & \left( (1-t)y+ t \frac{y}{\|y\|} ,(1-t)z \right) \end{array} \right..$$ Then $h \circ f = \mathrm{Id}$ and $H$ defines a homotopy from $\mathrm{Id}$ to $f \circ h$. In particular, $E \backslash Z$ and $S^{m-1}$ are homotopy equivalent hence $H_{m-1}(E \backslash Z \simeq H_{m-1}(S^{m-1})$.
In order to visualize what happens, consider for instance the situation where $Y= \mathbb{R}^2$ and $Z=\mathbb{R}$. Then $E \backslash Z= \{ (0,0,z) \in \mathbb{R}^3 \mid z \in \mathbb{R} \}$ is clearly homotopic equivalent to the unit circle $\{(x,y,0) \in \mathbb{R}^3 \mid x^2+y^2=1 \}$.
