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.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          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

          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


















          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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more