Characteristic map of a n-cell in a CW complex












6












$begingroup$


I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following:
Let $sigma$ be a n-cell and $Phi_sigma:mathbb D^n to X$ be the characteristic map. Is there any relation between $Phi_sigma(mathbb S^{n-1})$ and $partial sigma$ following directly from the Definition? I know, for example, that $Phi_sigma(mathbb S^{n-1})subset X^{n-1}$, $X^{n-1}$ being the $(n-1)$-Skeleton, but what does this tell me about the boundary of $sigma$?






general-topology algebraic-topology cw-complexes







share|cite|improve this question









$endgroup$

















    6












    $begingroup$


    I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following:
    Let $sigma$ be a n-cell and $Phi_sigma:mathbb D^n to X$ be the characteristic map. Is there any relation between $Phi_sigma(mathbb S^{n-1})$ and $partial sigma$ following directly from the Definition? I know, for example, that $Phi_sigma(mathbb S^{n-1})subset X^{n-1}$, $X^{n-1}$ being the $(n-1)$-Skeleton, but what does this tell me about the boundary of $sigma$?






    general-topology algebraic-topology cw-complexes







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      4



      $begingroup$


      I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following:
      Let $sigma$ be a n-cell and $Phi_sigma:mathbb D^n to X$ be the characteristic map. Is there any relation between $Phi_sigma(mathbb S^{n-1})$ and $partial sigma$ following directly from the Definition? I know, for example, that $Phi_sigma(mathbb S^{n-1})subset X^{n-1}$, $X^{n-1}$ being the $(n-1)$-Skeleton, but what does this tell me about the boundary of $sigma$?






      general-topology algebraic-topology cw-complexes







      share|cite|improve this question









      $endgroup$




      I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following:
      Let $sigma$ be a n-cell and $Phi_sigma:mathbb D^n to X$ be the characteristic map. Is there any relation between $Phi_sigma(mathbb S^{n-1})$ and $partial sigma$ following directly from the Definition? I know, for example, that $Phi_sigma(mathbb S^{n-1})subset X^{n-1}$, $X^{n-1}$ being the $(n-1)$-Skeleton, but what does this tell me about the boundary of $sigma$?






      general-topology algebraic-topology cw-complexes




      general-topology algebraic-topology cw-complexes






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 26 '13 at 14:56









      user83496user83496

      15310




      15310






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          If the cell $e^n_alpha$ is defined as $Phi_alpha(text{int}D^n)$, then you have the following relation:

          Since the characteristic map $Phi_alpha$ is continuous, you always have $Phi_alpha(D^n)=Phi_alphaleft(overline{text{int}D^n}right)subseteqoverline{e^n_alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $Phi(D^n)$ is closed, and since it contains $e^n_alpha$, it also contains the closure $overline{e^n_alpha}$. So in the end, $Phi_alpha(D^n)=overline{e^n_alpha}$.



          But beware that $partial e_alpha$ is in general not the same as $Phi_alphaleft(S^{n-1}right)$. For example, $partial e_alpha$ can be the cell $e_alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
            $endgroup$
            – user83496
            Aug 26 '13 at 15:58












          • $begingroup$
            Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
            $endgroup$
            – Stefan Hamcke
            Aug 26 '13 at 16:15












          • $begingroup$
            @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
            $endgroup$
            – Zuriel
            Oct 28 '14 at 12:48










          • $begingroup$
            @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
            $endgroup$
            – Stefan Hamcke
            Dec 4 '14 at 15:15



















          4












          $begingroup$

          I think it helps to have some pictures. Here is a picture of a $1$-dimensional cell complex.



          face



          For a $2$-dimensional example, try attaching a $2$-cell to the unit interval $[0,1]$, itself thought of as a cell complex with two $0$ cells and one $1$-cell. (The mouth in the face.) Here are two examples of attaching a $2$-cell.



          two



          In the first the attaching map takes the boundary of the $2$-disc to the point $1$. In the second, it takes the boundary of the $2$-disc to somewhere in the middle of the unit interval.



          These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $X$ to a space $B$ by means of a map $f:A to B$ on a closed subspace $A$ of $X$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
            $endgroup$
            – Baby Dragon
            Aug 26 '13 at 18:10











          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          If the cell $e^n_alpha$ is defined as $Phi_alpha(text{int}D^n)$, then you have the following relation:

          Since the characteristic map $Phi_alpha$ is continuous, you always have $Phi_alpha(D^n)=Phi_alphaleft(overline{text{int}D^n}right)subseteqoverline{e^n_alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $Phi(D^n)$ is closed, and since it contains $e^n_alpha$, it also contains the closure $overline{e^n_alpha}$. So in the end, $Phi_alpha(D^n)=overline{e^n_alpha}$.



          But beware that $partial e_alpha$ is in general not the same as $Phi_alphaleft(S^{n-1}right)$. For example, $partial e_alpha$ can be the cell $e_alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
            $endgroup$
            – user83496
            Aug 26 '13 at 15:58












          • $begingroup$
            Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
            $endgroup$
            – Stefan Hamcke
            Aug 26 '13 at 16:15












          • $begingroup$
            @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
            $endgroup$
            – Zuriel
            Oct 28 '14 at 12:48










          • $begingroup$
            @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
            $endgroup$
            – Stefan Hamcke
            Dec 4 '14 at 15:15
















          3












          $begingroup$

          If the cell $e^n_alpha$ is defined as $Phi_alpha(text{int}D^n)$, then you have the following relation:

          Since the characteristic map $Phi_alpha$ is continuous, you always have $Phi_alpha(D^n)=Phi_alphaleft(overline{text{int}D^n}right)subseteqoverline{e^n_alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $Phi(D^n)$ is closed, and since it contains $e^n_alpha$, it also contains the closure $overline{e^n_alpha}$. So in the end, $Phi_alpha(D^n)=overline{e^n_alpha}$.



          But beware that $partial e_alpha$ is in general not the same as $Phi_alphaleft(S^{n-1}right)$. For example, $partial e_alpha$ can be the cell $e_alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
            $endgroup$
            – user83496
            Aug 26 '13 at 15:58












          • $begingroup$
            Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
            $endgroup$
            – Stefan Hamcke
            Aug 26 '13 at 16:15












          • $begingroup$
            @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
            $endgroup$
            – Zuriel
            Oct 28 '14 at 12:48










          • $begingroup$
            @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
            $endgroup$
            – Stefan Hamcke
            Dec 4 '14 at 15:15














          3












          3








          3





          $begingroup$

          If the cell $e^n_alpha$ is defined as $Phi_alpha(text{int}D^n)$, then you have the following relation:

          Since the characteristic map $Phi_alpha$ is continuous, you always have $Phi_alpha(D^n)=Phi_alphaleft(overline{text{int}D^n}right)subseteqoverline{e^n_alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $Phi(D^n)$ is closed, and since it contains $e^n_alpha$, it also contains the closure $overline{e^n_alpha}$. So in the end, $Phi_alpha(D^n)=overline{e^n_alpha}$.



          But beware that $partial e_alpha$ is in general not the same as $Phi_alphaleft(S^{n-1}right)$. For example, $partial e_alpha$ can be the cell $e_alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.






          share|cite|improve this answer











          $endgroup$



          If the cell $e^n_alpha$ is defined as $Phi_alpha(text{int}D^n)$, then you have the following relation:

          Since the characteristic map $Phi_alpha$ is continuous, you always have $Phi_alpha(D^n)=Phi_alphaleft(overline{text{int}D^n}right)subseteqoverline{e^n_alpha}$. On the other hand, once you have proven that CW-complexes are Hausdorff (and, more generally, normal), you know that compact subsets are closed. Therefore $Phi(D^n)$ is closed, and since it contains $e^n_alpha$, it also contains the closure $overline{e^n_alpha}$. So in the end, $Phi_alpha(D^n)=overline{e^n_alpha}$.



          But beware that $partial e_alpha$ is in general not the same as $Phi_alphaleft(S^{n-1}right)$. For example, $partial e_alpha$ can be the cell $e_alpha$ itself if it's a $2$-cell with a $3$-cell glued to it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '14 at 15:14

























          answered Aug 26 '13 at 15:17









          Stefan HamckeStefan Hamcke

          21.7k42879




          21.7k42879












          • $begingroup$
            Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
            $endgroup$
            – user83496
            Aug 26 '13 at 15:58












          • $begingroup$
            Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
            $endgroup$
            – Stefan Hamcke
            Aug 26 '13 at 16:15












          • $begingroup$
            @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
            $endgroup$
            – Zuriel
            Oct 28 '14 at 12:48










          • $begingroup$
            @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
            $endgroup$
            – Stefan Hamcke
            Dec 4 '14 at 15:15


















          • $begingroup$
            Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
            $endgroup$
            – user83496
            Aug 26 '13 at 15:58












          • $begingroup$
            Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
            $endgroup$
            – Stefan Hamcke
            Aug 26 '13 at 16:15












          • $begingroup$
            @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
            $endgroup$
            – Zuriel
            Oct 28 '14 at 12:48










          • $begingroup$
            @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
            $endgroup$
            – Stefan Hamcke
            Dec 4 '14 at 15:15
















          $begingroup$
          Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
          $endgroup$
          – user83496
          Aug 26 '13 at 15:58






          $begingroup$
          Of course, thx! So, at least $overline{e^n_alpha}e^n_alpha$ is contained in $X^{n-1}$, that's very encouraging :)
          $endgroup$
          – user83496
          Aug 26 '13 at 15:58














          $begingroup$
          Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
          $endgroup$
          – Stefan Hamcke
          Aug 26 '13 at 16:15






          $begingroup$
          Yes, because $overline{e^n_alpha}-e^n_alpha$ is the same as $Phi_alpha(S^{n-1})$ since $Phi(partial D^n)$ and $e^n_alpha$ are disjoint.
          $endgroup$
          – Stefan Hamcke
          Aug 26 '13 at 16:15














          $begingroup$
          @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
          $endgroup$
          – Zuriel
          Oct 28 '14 at 12:48




          $begingroup$
          @StefanHamcke, I am reading Introduction to Homotopy Theory by Martin Arkowitz. Exercise 1.16 of the book asks the reader to prove that the boundary of $e_{alpha}$ is equal to $Phi(S^{n-1})$. Does it contradict the first sentence of the second paragraph of your answer?
          $endgroup$
          – Zuriel
          Oct 28 '14 at 12:48












          $begingroup$
          @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
          $endgroup$
          – Stefan Hamcke
          Dec 4 '14 at 15:15




          $begingroup$
          @Zuriel: Sorry for the late answer, I had to finish my diploma thesis. The boundary of a cell $e_alpha$ must contain $overline{e_α}-e_α=Phileft(S^{n-1}right)$, but it could be all of $e_α$, for example when $e_α$ is the $1$-cell in a complex consisting of a $1$-cell whose end points are attached to a point, and a $2$-cell which is a disk attached along its boundary via a homeomorphism to the circle $overline{e_α}$.
          $endgroup$
          – Stefan Hamcke
          Dec 4 '14 at 15:15











          4












          $begingroup$

          I think it helps to have some pictures. Here is a picture of a $1$-dimensional cell complex.



          face



          For a $2$-dimensional example, try attaching a $2$-cell to the unit interval $[0,1]$, itself thought of as a cell complex with two $0$ cells and one $1$-cell. (The mouth in the face.) Here are two examples of attaching a $2$-cell.



          two



          In the first the attaching map takes the boundary of the $2$-disc to the point $1$. In the second, it takes the boundary of the $2$-disc to somewhere in the middle of the unit interval.



          These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $X$ to a space $B$ by means of a map $f:A to B$ on a closed subspace $A$ of $X$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
            $endgroup$
            – Baby Dragon
            Aug 26 '13 at 18:10
















          4












          $begingroup$

          I think it helps to have some pictures. Here is a picture of a $1$-dimensional cell complex.



          face



          For a $2$-dimensional example, try attaching a $2$-cell to the unit interval $[0,1]$, itself thought of as a cell complex with two $0$ cells and one $1$-cell. (The mouth in the face.) Here are two examples of attaching a $2$-cell.



          two



          In the first the attaching map takes the boundary of the $2$-disc to the point $1$. In the second, it takes the boundary of the $2$-disc to somewhere in the middle of the unit interval.



          These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $X$ to a space $B$ by means of a map $f:A to B$ on a closed subspace $A$ of $X$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
            $endgroup$
            – Baby Dragon
            Aug 26 '13 at 18:10














          4












          4








          4





          $begingroup$

          I think it helps to have some pictures. Here is a picture of a $1$-dimensional cell complex.



          face



          For a $2$-dimensional example, try attaching a $2$-cell to the unit interval $[0,1]$, itself thought of as a cell complex with two $0$ cells and one $1$-cell. (The mouth in the face.) Here are two examples of attaching a $2$-cell.



          two



          In the first the attaching map takes the boundary of the $2$-disc to the point $1$. In the second, it takes the boundary of the $2$-disc to somewhere in the middle of the unit interval.



          These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $X$ to a space $B$ by means of a map $f:A to B$ on a closed subspace $A$ of $X$.






          share|cite|improve this answer











          $endgroup$



          I think it helps to have some pictures. Here is a picture of a $1$-dimensional cell complex.



          face



          For a $2$-dimensional example, try attaching a $2$-cell to the unit interval $[0,1]$, itself thought of as a cell complex with two $0$ cells and one $1$-cell. (The mouth in the face.) Here are two examples of attaching a $2$-cell.



          two



          In the first the attaching map takes the boundary of the $2$-disc to the point $1$. In the second, it takes the boundary of the $2$-disc to somewhere in the middle of the unit interval.



          These pictures are taken from Topology and Groupoids (colour in the e-version). I think it helps also to develop some of the general theory of attaching a space $X$ to a space $B$ by means of a map $f:A to B$ on a closed subspace $A$ of $X$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 21:14









          Glorfindel

          3,41981830




          3,41981830










          answered Aug 26 '13 at 18:01









          Ronnie BrownRonnie Brown

          12.1k12939




          12.1k12939








          • 1




            $begingroup$
            I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
            $endgroup$
            – Baby Dragon
            Aug 26 '13 at 18:10














          • 1




            $begingroup$
            I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
            $endgroup$
            – Baby Dragon
            Aug 26 '13 at 18:10








          1




          1




          $begingroup$
          I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
          $endgroup$
          – Baby Dragon
          Aug 26 '13 at 18:10




          $begingroup$
          I think that you should name him Alphalpha. uploads.neatorama.com/wp-content/uploads/2010/07/…
          $endgroup$
          – Baby Dragon
          Aug 26 '13 at 18:10


















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