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








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f476582%2fcharacteristic-map-of-a-n-cell-in-a-cw-complex%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          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

          22k42880




          22k42880












          • $begingroup$
            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

          3,41381930




          3,41381930










          answered Aug 26 '13 at 18:01









          Ronnie BrownRonnie Brown

          12.2k13039




          12.2k13039








          • 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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f476582%2fcharacteristic-map-of-a-n-cell-in-a-cw-complex%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
          Read more

          Puebla de Zaragoza

          Image
          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
          Read more

          Musa

          Image
          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
          Read more