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I have the following problem that I am stuck on.

Let $A$ be a subspace of $X$ and let $Y$ be a non-empty topological space. Show that $A\times Y$ is a retract of $X\times Y$ if and only if $A$ is a retract of $X$.

My work so far:

$(\Rightarrow)$ Assume that $A\times Y$ is a retract of $X\times Y$. Then there exists a continuous map $r:X\times Y\to A\times Y$ such that $r|_{A\times Y}=\mathrm{id}_{A\times Y}$. In other words, $r(a,y)=(a,y)$ for all $(a,y)\in A\times Y$. Then $r=\mathrm{id}_{A}\times\mathrm{id}_{Y}$, so $A$ is a retract of $X$ through the map $\mathrm{id}_{A}$.

I don't think any of this is right, and I am super confused in how to show the result. Thanks in advance for any help!

John Griffin
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Sir_Math_Cat
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1 Answers1

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Suppose $A\times Y$ is a retract of $X\times Y$. We want to show that $A$ is a retract, which means finding a continuous function $f:X\to A$ such that $f|_A=\operatorname{id}_A$. Since $A\times Y$ is a retract of $X\times Y$, we have a continuous function $r:X\times Y\to A\times Y$ such that $r|_{X\times Y}=\operatorname{id}_{A\times Y}$. Since $Y$ is nonempty, we can find $y_0\in Y$. Then define $f(x)=\pi_1(r(x,y_0))$ for every $x\in X$, where $\pi_1:X\times Y\to X$ denotes the projection. Verify that $f$ is continuous and that it is a retraction.

Conversely, suppose $A$ is a retract of $X$. Then there exists a retraction $r:X\to A$. Can you think of a way to define a retraction $f:X\times Y\to A\times Y$? Let me know if you get stuck and I'll add more.

John Griffin
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  • The forward direction makes so much more sense now; thank you! I'm still not quite sure about the converses, though. – Sir_Math_Cat Oct 16 '17 at 01:31
  • @WarTurtle It's the same idea. You need to use the retraction given to define the retraction desired. Given an element in $(x,y)\in X\times Y$, how do we get an element in $A\times Y$ using the fact that we can use $r$ to map elements from $X$ into $A$? – John Griffin Oct 16 '17 at 01:33
  • Do we use the inverse of the projection map? – Sir_Math_Cat Oct 16 '17 at 01:34
  • @WarTurtle No - projection maps aren't invertible. Try combining the map $r:X\to Y$ with the identity on $Y$. – John Griffin Oct 16 '17 at 01:40
  • So we would consider the map, say, $g=r\times \mathrm{id}{y}=(r(x),\mathrm{id}{Y}(y))$? – Sir_Math_Cat Oct 16 '17 at 01:44
  • @WarTurtle Maybe. You tell me. Does that work? It it continuous and is it the identity on $A\times Y$? – John Griffin Oct 16 '17 at 01:44
  • I think? I'm not sure how to actually show it. – Sir_Math_Cat Oct 16 '17 at 01:46
  • @WarTurtle Showing it's the identity on $A\times Y$ is simple: We have $$g(a,y)=(r(a),\operatorname{id}_Y(y))=(a,y)$$ because $r(a)=a$ for every $a\in A$. There are many ways to show continuity. What are the ways you know to show continuity of a function? – John Griffin Oct 16 '17 at 01:52
  • I just need to show that it takes open sets to open sets, right? – Sir_Math_Cat Oct 16 '17 at 01:53
  • @WarTurtle A map that takes open sets to open sets is not necessarily continuous. The definition of continuous is that the preimage of an open set is open. Alternatively you can use this: https://proofwiki.org/wiki/Function_to_Product_Space_is_Continuous_iff_Composition_with_Projections_are_Continuous. Yet again, if you've seen nets before, you can show this "version" of continuity: https://math.stackexchange.com/questions/360419/f-brings-convergent-nets-to-convergent-nets-is-it-continuous (see Pete's answer). – John Griffin Oct 16 '17 at 01:58
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    Ah, yes, I mixed up my definition of continuity and open maps. Thank you for all your help! – Sir_Math_Cat Oct 16 '17 at 01:59