# Sufficient condition for Existence of continous retraction

• Apr 23rd 2010, 01:20 PM
Xian
Sufficient condition for Existence of continous retraction
If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

If so, how can this be proven?
If not, any counterexamples?

Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.
• Apr 23rd 2010, 02:35 PM
Drexel28
Quote:

Originally Posted by Xian
If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

If so, how can this be proven?
If not, any counterexamples?

Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.

This question makes no sense. What are you asking?
• Apr 23rd 2010, 03:07 PM
Xian
Quote:

Originally Posted by Xian
If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

If so, how can this be proven?
If not, any counterexamples?

Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.

What I mean is that the the existence of continuous map $f:X\rightarrow X$ where $f(X)=A\subseteq X$ should guarantee the existence of a continuous retraction of $X$ to $A$.

Actually I'd like to make my statement even stronger:

If there is a function $f:X\rightarrow X$ where $f(X)\subseteq A\subseteq X$ then there exists a continuous retraction $r:X\rightarrow A$ such that $\forall a\in A$ $r(a)=a$.

Is this true? Proof?
Is it false? Counterexample?
• Apr 23rd 2010, 03:11 PM
Drexel28
Quote:

Originally Posted by Xian
What I mean is that the the existence of continuous map $f:X\rightarrow X$ where $f(X)=A\subseteq X$ should guarantee the existence of a continuous retraction of $X$ to $A$.

Actually I'd like to make my statement even stronger:

If there is a function $f:X\rightarrow X$ where $f(X)\subseteq A\subseteq X$ then there exists a continuous retraction $r:X\rightarrow A$ such that $\forall a\in A$ $r(a)=a$.

Is this true? Proof?
Is it false? Counterexample?

I just am still lost. Why do we have two maps? A deformation retract is a map $r:X\to A\subseteq X$ such that $r\circ\iota_A=\text{id}_A$. What does $f$ have to do with this? Are you discussing sufficient conditions for $f$ to bea deformation retract?
• Apr 23rd 2010, 09:23 PM
Xian
Quote:

Originally Posted by Drexel28
I just am still lost. Why do we have two maps? A deformation retract is a map $r:X\to A\subseteq X$ such that $r\circ\iota_A=\text{id}_A$. What does $f$ have to do with this? Are you discussing sufficient conditions for $f$ to bea deformation retract?

I'm saying suppose such an f exists with the above properties. Does the existence of f imply the existence of r?