Example Of A Non-Existent Retract












0














I am looking for an example that disproves the claim that given any subspace $A$ of a topological space $X$, there exists a retract of $X$ onto $A$.






general-topology algebraic-topology







share|cite|improve this question


















  • 3




    $S^1 subset mathbb{R}^2$
    – Tim kinsella
    Nov 26 at 1:41






  • 1




    How about the closed unit disk and the unit circle?
    – ncmathsadist
    Nov 26 at 1:41










  • Oh, duh, of course. Silly me. Thank you.
    – Frederic Chopin
    Nov 26 at 1:43










  • not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
    – Tim kinsella
    Nov 26 at 2:07










  • That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
    – Frederic Chopin
    Nov 26 at 2:11


















0














I am looking for an example that disproves the claim that given any subspace $A$ of a topological space $X$, there exists a retract of $X$ onto $A$.






general-topology algebraic-topology







share|cite|improve this question


















  • 3




    $S^1 subset mathbb{R}^2$
    – Tim kinsella
    Nov 26 at 1:41






  • 1




    How about the closed unit disk and the unit circle?
    – ncmathsadist
    Nov 26 at 1:41










  • Oh, duh, of course. Silly me. Thank you.
    – Frederic Chopin
    Nov 26 at 1:43










  • not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
    – Tim kinsella
    Nov 26 at 2:07










  • That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
    – Frederic Chopin
    Nov 26 at 2:11
















0












0








0


1





I am looking for an example that disproves the claim that given any subspace $A$ of a topological space $X$, there exists a retract of $X$ onto $A$.






general-topology algebraic-topology







share|cite|improve this question













I am looking for an example that disproves the claim that given any subspace $A$ of a topological space $X$, there exists a retract of $X$ onto $A$.






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 1:39









Frederic Chopin

322111




322111








  • 3




    $S^1 subset mathbb{R}^2$
    – Tim kinsella
    Nov 26 at 1:41






  • 1




    How about the closed unit disk and the unit circle?
    – ncmathsadist
    Nov 26 at 1:41










  • Oh, duh, of course. Silly me. Thank you.
    – Frederic Chopin
    Nov 26 at 1:43










  • not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
    – Tim kinsella
    Nov 26 at 2:07










  • That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
    – Frederic Chopin
    Nov 26 at 2:11
















  • 3




    $S^1 subset mathbb{R}^2$
    – Tim kinsella
    Nov 26 at 1:41






  • 1




    How about the closed unit disk and the unit circle?
    – ncmathsadist
    Nov 26 at 1:41










  • Oh, duh, of course. Silly me. Thank you.
    – Frederic Chopin
    Nov 26 at 1:43










  • not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
    – Tim kinsella
    Nov 26 at 2:07










  • That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
    – Frederic Chopin
    Nov 26 at 2:11










3




3




$S^1 subset mathbb{R}^2$
– Tim kinsella
Nov 26 at 1:41




$S^1 subset mathbb{R}^2$
– Tim kinsella
Nov 26 at 1:41




1




1




How about the closed unit disk and the unit circle?
– ncmathsadist
Nov 26 at 1:41




How about the closed unit disk and the unit circle?
– ncmathsadist
Nov 26 at 1:41












Oh, duh, of course. Silly me. Thank you.
– Frederic Chopin
Nov 26 at 1:43




Oh, duh, of course. Silly me. Thank you.
– Frederic Chopin
Nov 26 at 1:43












not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
– Tim kinsella
Nov 26 at 2:07




not silly -- its not at all obvious unless you think about it in terms of $pi_1$. $X$ retracts onto $A$ only if $pi_1(A)$ is a subgroup of $pi_1(X)$.
– Tim kinsella
Nov 26 at 2:07












That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
– Frederic Chopin
Nov 26 at 2:11






That's true. But I was thinking indeed about it in terms of the fundamental group, so recalling that the homomorphism induced by the inclusion map is an injection if the space retracts onto some subspace makes this example clear immediately.
– Frederic Chopin
Nov 26 at 2:11












2 Answers
2






active

oldest

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5














Here's the simplest possible example:



Consider the space $X$ with three points $a,b,c$ and open sets $$emptyset, {a},{c}, {a,c}, {a,b,c}.$$ Let $A={a,c}$. There are only two maps from $X$ to $A$ which are the identity on $A$, and neither is continuous. E.g. if we send $b$ to $a$, then the preimage of the open set ${a}$ is the non-open set ${a,b}$.



This really is the simplest example, since any space retracts onto any of its singleton subsets and onto itself.






share|cite|improve this answer



















  • 2




    This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
    – Alex Kruckman
    Nov 26 at 2:52












  • @AlexKruckman Thanks, and quite right - fixed!
    – Noah Schweber
    Nov 26 at 3:08










  • @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
    – Tim kinsella
    Nov 29 at 6:43



















2














A retract of a Hausdorff space is closed so $(0,1)$ is not a retract of $mathbb{R}$ (usual topology).






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    Here's the simplest possible example:



    Consider the space $X$ with three points $a,b,c$ and open sets $$emptyset, {a},{c}, {a,c}, {a,b,c}.$$ Let $A={a,c}$. There are only two maps from $X$ to $A$ which are the identity on $A$, and neither is continuous. E.g. if we send $b$ to $a$, then the preimage of the open set ${a}$ is the non-open set ${a,b}$.



    This really is the simplest example, since any space retracts onto any of its singleton subsets and onto itself.






    share|cite|improve this answer



















    • 2




      This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
      – Alex Kruckman
      Nov 26 at 2:52












    • @AlexKruckman Thanks, and quite right - fixed!
      – Noah Schweber
      Nov 26 at 3:08










    • @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
      – Tim kinsella
      Nov 29 at 6:43
















    5














    Here's the simplest possible example:



    Consider the space $X$ with three points $a,b,c$ and open sets $$emptyset, {a},{c}, {a,c}, {a,b,c}.$$ Let $A={a,c}$. There are only two maps from $X$ to $A$ which are the identity on $A$, and neither is continuous. E.g. if we send $b$ to $a$, then the preimage of the open set ${a}$ is the non-open set ${a,b}$.



    This really is the simplest example, since any space retracts onto any of its singleton subsets and onto itself.






    share|cite|improve this answer



















    • 2




      This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
      – Alex Kruckman
      Nov 26 at 2:52












    • @AlexKruckman Thanks, and quite right - fixed!
      – Noah Schweber
      Nov 26 at 3:08










    • @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
      – Tim kinsella
      Nov 29 at 6:43














    5












    5








    5






    Here's the simplest possible example:



    Consider the space $X$ with three points $a,b,c$ and open sets $$emptyset, {a},{c}, {a,c}, {a,b,c}.$$ Let $A={a,c}$. There are only two maps from $X$ to $A$ which are the identity on $A$, and neither is continuous. E.g. if we send $b$ to $a$, then the preimage of the open set ${a}$ is the non-open set ${a,b}$.



    This really is the simplest example, since any space retracts onto any of its singleton subsets and onto itself.






    share|cite|improve this answer














    Here's the simplest possible example:



    Consider the space $X$ with three points $a,b,c$ and open sets $$emptyset, {a},{c}, {a,c}, {a,b,c}.$$ Let $A={a,c}$. There are only two maps from $X$ to $A$ which are the identity on $A$, and neither is continuous. E.g. if we send $b$ to $a$, then the preimage of the open set ${a}$ is the non-open set ${a,b}$.



    This really is the simplest example, since any space retracts onto any of its singleton subsets and onto itself.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 at 3:08

























    answered Nov 26 at 2:03









    Noah Schweber

    120k10146279




    120k10146279








    • 2




      This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
      – Alex Kruckman
      Nov 26 at 2:52












    • @AlexKruckman Thanks, and quite right - fixed!
      – Noah Schweber
      Nov 26 at 3:08










    • @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
      – Tim kinsella
      Nov 29 at 6:43














    • 2




      This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
      – Alex Kruckman
      Nov 26 at 2:52












    • @AlexKruckman Thanks, and quite right - fixed!
      – Noah Schweber
      Nov 26 at 3:08










    • @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
      – Tim kinsella
      Nov 29 at 6:43








    2




    2




    This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
    – Alex Kruckman
    Nov 26 at 2:52






    This is my favorite kind of answer to general topology questions, +1! Why get $pi_1$ involved if you don't have to? But I think $A$ should be ${a,c}$.
    – Alex Kruckman
    Nov 26 at 2:52














    @AlexKruckman Thanks, and quite right - fixed!
    – Noah Schweber
    Nov 26 at 3:08




    @AlexKruckman Thanks, and quite right - fixed!
    – Noah Schweber
    Nov 26 at 3:08












    @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
    – Tim kinsella
    Nov 29 at 6:43




    @AlexKruckman yeah, this is a better answer. but note that this is just using $pi_0$ instead of $pi_1$.
    – Tim kinsella
    Nov 29 at 6:43











    2














    A retract of a Hausdorff space is closed so $(0,1)$ is not a retract of $mathbb{R}$ (usual topology).






    share|cite|improve this answer


























      2














      A retract of a Hausdorff space is closed so $(0,1)$ is not a retract of $mathbb{R}$ (usual topology).






      share|cite|improve this answer
























        2












        2








        2






        A retract of a Hausdorff space is closed so $(0,1)$ is not a retract of $mathbb{R}$ (usual topology).






        share|cite|improve this answer












        A retract of a Hausdorff space is closed so $(0,1)$ is not a retract of $mathbb{R}$ (usual topology).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 6:17









        Henno Brandsma

        104k346113




        104k346113






























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