Consider a separable, infinite dimesnional Hilbert space $H$. I want to show that the inclusion into the odd indices $(e_1,e_2,e_3,...)\mapsto (e_1,0,e_2,0,e_3,0,...)$ is isometrically homotopic to the identity. I know that the homotopy $(0,e)\mapsto (\mathrm{sin}(t\pi/2)e,cos(t\pi/2)e)$ rotates the second component to the first and therefore yields an isometric homotopy between the inclusion to the odd indices to the inclusion $(e_1,e_2,e_3,...)\mapsto (e_1,e_2,0,0,e_3,0,e_4,0,...)$. Unfortunately one can not do this for all components at once since then we obtain mixed terms which fail to be isometric. Maybe it could be possible to argue that we can do this subsequently for each component. Can somebody help me with this argument?
I think that the countable concatenation of homotopies is not possible in general.
It should be possible to show that inductively rotating the components converges pointwise to the indentity, because for a fixed element $e\in H$ there is for each $\epsilon>0$ an $n_0\in\mathbb{N}$ such that the norm of $(0,...,0,e_{n_0},e_{n_0+1},...)$ is smaller than $\epsilon$. Is this enough?