# Thread: Sufficient condition for Existence of continous retraction

1. ## 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.

2. 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?

3. 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 $\displaystyle f:X\rightarrow X$ where $\displaystyle f(X)=A\subseteq X$ should guarantee the existence of a continuous retraction of $\displaystyle X$ to $\displaystyle A$.

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

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

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

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

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

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

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 $\displaystyle r:X\to A\subseteq X$ such that $\displaystyle r\circ\iota_A=\text{id}_A$. What does $\displaystyle f$ have to do with this? Are you discussing sufficient conditions for $\displaystyle f$ to bea deformation retract?

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