A $X_0$ path-component of $X$ is as retract of $X$?

Question

1. A $X_0$ path-component of $X_0$ is as retract of $X$ ?
2. $X_0$ is a deformation retract of $X$?

There is some counter exemple for this?

I am think in a exemple:

Let $X$ be a union of two spheres with the same origin and distinc radius and $X_0$ be the path-component of $X$. $X_0$ is a deformation retract of $X$? $X$ is not path connected, but the path-component of $X$ is some of this two spheres ( by definition of path-component, $X_0$ must be the sphere with biggest radius). There is a deformation retraction of $X_0$ onto $X$?

Answer 1

Both a retract and deformation retract of a Hausdorff space has to be closed, see this: Show that a retract of a Hausdorff space is closed.

So all you need is to find a Hausdorff space $X$ with a path component that is not closed. One such example is the closure of topologists sine curve:

$$T=\big\{\big(x,\sin(1/x)\big)\ \big|\ x>0\big\}$$ $$X=\big(\{0\}\times [-1,1]\big)\cup T$$

and then $T$ is your path component that is not a (deformation) retract.

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Your example doesn't really work. It is indeed the case that $X_0$ is not a deformation retract of $X$, because deformation retract has to have the same number of path components. That's because deformation retraction is a special case of homotopy equivalence. In particular a path component $X_0$ is deformation retract of $X$ only when $X$ is path connected, i.e. $X_0=X$.

However your $X_0$ is a retract of $X$. If you normalize your example, so that $X_0$ is a unit sphere, then the retraction is given by normalization $x\mapsto \frac{x}{\lVert x\rVert}$.

Answer 2

Consider the topologist's sine curve, here there's a path component that is not closed. But in a Hausdorff space, retracts are always closed. Thus there exist an $X$ such that $X_0$ is not a retract of $X$. As every deformation retract is in particular a retract, this answers the second question as well. (But the second question is much easier: any space which is not path connected works)