Retract of a connected space is connected

[PsychonautQQ]

let r: X-->A be a retraction. This means that r*i = Id_A, where * is composition of maps and i is the inclusion map from A--->X.

How to show that the retract of a connected space is also connected? I was trying to proceed by contradiction.

Suppose A is not connected, then A = Union(M,N) where M and N are open in A and Intersection(M,N) = empty.

Then i used the inclusion map on A, ie:
i: A---> X
= i: Union(M,N) --> X

Where M = Intersection(A,R) for R open in X and N=Intersection(A,S) for S open in X, so:

= i: Union(intersection(A,R),Intersection(A,S)) --- > X

Perhaps I can show that since this is an embedding, that using the image of i in X I can find a separation of X. I don't know if A is open or closed in X though, which might be useful information.

I'm sure there is a much easier way to do this.

Thanks PF!

 

[andrewkirk]

I'm sure there is a much easier way to do this.

There is
(At least I think so)

Start as you started your proof, with $A=M\cup N$ and $N,M$ open in $A$. We then use the crucial property of a retraction that you omitted to mention, which is that it is continuous. What does that enable us to say about $r^{-1}(M)$ and $r^{-1}(N)$ that would contradict the assumption that $X$ is connected?

We didn't need to use the inclusion map at all, so the result must be true for any continuous map from a connected space to a subspace of itself.

 

[PsychonautQQ]

There is
(At least I think so)

Start as you started your proof, with $A=M\cup N$ and $N,M$ open in $A$. We then use the crucial property of a retraction that you omitted to mention, which is that it is continuous. What does that enable us to say about $r^{-1}(M)$ and $r^{-1}(N)$ that would contradict the assumption that $X$ is connected?

We didn't need to use the inclusion map at all, so the result must be true for any continuous map from a connected space to a subspace of itself.

$r^{-1}(M)$ and $r^{-1}(N)$ will be open in X. And since $r$ surjects $X$ onto $A$, it must be that union of these two pre-images cover all of X. I'm assuming that the intersection of these images is going to also be disjoint since the map is continuous?

 

[andrewkirk]

I'm assuming that the intersection of these images is going to also be disjoint since the map is continuous?

We do not need continuity for that. It follows from the fact that r is a function and that M, N are disjoint (which latter I omitted to state above, but is implied by M, N being a separation of A).

 

[WWGD]

If you want the big machinery, then a retraction is a homotopy equivalence, and homotopy equivalence preserves connectedness ( since, e.g., it preserves homology)

 

[Infrared]

If you want the big machinery, then a retraction is a homotopy equivalence, and homotopy equivalence preserves connectedness ( since, e.g., it preserves homology)

A retraction is not necessarily a homotopy equivalence. Any nonempty space retracts onto a point.

 

[WWGD]

A retraction is not necessarily a homotopy equivalence. Any nonempty space retracts onto a point.

My bad, I meant deformation retract.