Are locally contractible spaces hereditarily paracompact?












2












$begingroup$


The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.



I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $mathcal F(U) to mathcal F^+(U)$ is surjective for all open subsets $U subseteq X$, assuming the existence part of the glueing axiom.



So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.






general-topology algebraic-topology paracompactness







share|cite|improve this question











$endgroup$












  • $begingroup$
    I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 4:16












  • $begingroup$
    Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 23:09
















2












$begingroup$


The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.



I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $mathcal F(U) to mathcal F^+(U)$ is surjective for all open subsets $U subseteq X$, assuming the existence part of the glueing axiom.



So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.






general-topology algebraic-topology paracompactness







share|cite|improve this question











$endgroup$












  • $begingroup$
    I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 4:16












  • $begingroup$
    Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 23:09














2












2








2





$begingroup$


The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.



I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $mathcal F(U) to mathcal F^+(U)$ is surjective for all open subsets $U subseteq X$, assuming the existence part of the glueing axiom.



So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.






general-topology algebraic-topology paracompactness







share|cite|improve this question











$endgroup$




The question title says it all. For the record, I have no reason to believe that this is true, but my question has a bit of a background.



I am reading Ramanan's Global Calculus book because I am interested in the isomorphism between singular cohomology with coefficients in a ring $R$ and sheaf cohomology with respect to the constant sheaf $underline R$ (page 114, Theorem 4.14). He only assumes his topological space $X$ is locally contractible, but he quotes a result (page 8,Lemma 1.14) which assumes his space is hereditarily paracompact ; the result says the sheafification map of a presheaf $mathcal F(U) to mathcal F^+(U)$ is surjective for all open subsets $U subseteq X$, assuming the existence part of the glueing axiom.



So I was wondering if he just wrote somewhere he would make that assumption for a long time, forgot to assume it, or if the question in my title has a positive answer, which to be honest, would surprise me! But hey, until I have a counter-example... who knows.






general-topology algebraic-topology paracompactness




general-topology algebraic-topology paracompactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 3:33









Eric Wofsey

184k13212338




184k13212338










asked Apr 21 '16 at 17:46









Patrick Da SilvaPatrick Da Silva

32k354106




32k354106












  • $begingroup$
    I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 4:16












  • $begingroup$
    Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 23:09


















  • $begingroup$
    I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 4:16












  • $begingroup$
    Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
    $endgroup$
    – Eric Auld
    May 22 '16 at 23:09
















$begingroup$
I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
$endgroup$
– Eric Auld
May 22 '16 at 4:16






$begingroup$
I just asked a similar question for exactly the same reason! (Also because I am dissatisfied with some other proofs of the equivalence of singular/sheaf cohomology, which seem to rely on the surjectivity of the sheafification map, though they assume only local contractibility.) math.stackexchange.com/questions/1794725/…
$endgroup$
– Eric Auld
May 22 '16 at 4:16














$begingroup$
Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
$endgroup$
– Eric Auld
May 22 '16 at 23:09




$begingroup$
Also, I think I found a serious mistake in that chapter math.stackexchange.com/questions/1795745/…
$endgroup$
– Eric Auld
May 22 '16 at 23:09










3 Answers
3






active

oldest

votes


















2












$begingroup$

The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you have an idea about what happened in the book? Thanks for bringing up the example though!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:55












  • $begingroup$
    @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
    $endgroup$
    – Henno Brandsma
    Apr 22 '16 at 7:53










  • $begingroup$
    @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
    $endgroup$
    – Patrick Da Silva
    Apr 22 '16 at 12:04



















1












$begingroup$

A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $mathbb{R}$, but not paracompact.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Same question as the one I asked Captain Lama!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:56










  • $begingroup$
    The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
    $endgroup$
    – Eric Wofsey
    Apr 21 '16 at 17:59










  • $begingroup$
    @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 18:04





















1












$begingroup$

To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility,
in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3),
where he also explains how Ramanan uses the hereditary paracompactness condition.






share|cite|improve this answer









$endgroup$













    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%2f1753095%2fare-locally-contractible-spaces-hereditarily-paracompact%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Do you have an idea about what happened in the book? Thanks for bringing up the example though!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:55












    • $begingroup$
      @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
      $endgroup$
      – Henno Brandsma
      Apr 22 '16 at 7:53










    • $begingroup$
      @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
      $endgroup$
      – Patrick Da Silva
      Apr 22 '16 at 12:04
















    2












    $begingroup$

    The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Do you have an idea about what happened in the book? Thanks for bringing up the example though!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:55












    • $begingroup$
      @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
      $endgroup$
      – Henno Brandsma
      Apr 22 '16 at 7:53










    • $begingroup$
      @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
      $endgroup$
      – Patrick Da Silva
      Apr 22 '16 at 12:04














    2












    2








    2





    $begingroup$

    The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.






    share|cite|improve this answer









    $endgroup$



    The long line should be a counter-example : it is known that it's not paracompact (see https://en.wikipedia.org/wiki/Long_line_%28topology%29), but it's a manifold if we omit the "second countable" condition from the definition, so in particular it's locally homeomorphic to an open interval, and thus locally contractible.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 21 '16 at 17:54









    Captain LamaCaptain Lama

    10k829




    10k829












    • $begingroup$
      Do you have an idea about what happened in the book? Thanks for bringing up the example though!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:55












    • $begingroup$
      @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
      $endgroup$
      – Henno Brandsma
      Apr 22 '16 at 7:53










    • $begingroup$
      @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
      $endgroup$
      – Patrick Da Silva
      Apr 22 '16 at 12:04


















    • $begingroup$
      Do you have an idea about what happened in the book? Thanks for bringing up the example though!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:55












    • $begingroup$
      @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
      $endgroup$
      – Henno Brandsma
      Apr 22 '16 at 7:53










    • $begingroup$
      @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
      $endgroup$
      – Patrick Da Silva
      Apr 22 '16 at 12:04
















    $begingroup$
    Do you have an idea about what happened in the book? Thanks for bringing up the example though!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:55






    $begingroup$
    Do you have an idea about what happened in the book? Thanks for bringing up the example though!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:55














    $begingroup$
    @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
    $endgroup$
    – Henno Brandsma
    Apr 22 '16 at 7:53




    $begingroup$
    @PatrickDaSilva many books assume manifolds are second countable (and then they are metrisable). But this seems not to be the case here, or hereditarily paracompact would not turn up like this (all metric spaces have that property, so it doesn't add anything). So I'm likewise mystified.
    $endgroup$
    – Henno Brandsma
    Apr 22 '16 at 7:53












    $begingroup$
    @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
    $endgroup$
    – Patrick Da Silva
    Apr 22 '16 at 12:04




    $begingroup$
    @Henno Brandsma : Indeed, like I said I didn't believe that locally contractible implied hereditarily paracompact, but the book kind of put me in that situation. It clearly says "locally contractible topological space" and then refers to that result... what a pain.
    $endgroup$
    – Patrick Da Silva
    Apr 22 '16 at 12:04











    1












    $begingroup$

    A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $mathbb{R}$, but not paracompact.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Same question as the one I asked Captain Lama!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:56










    • $begingroup$
      The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
      $endgroup$
      – Eric Wofsey
      Apr 21 '16 at 17:59










    • $begingroup$
      @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 18:04


















    1












    $begingroup$

    A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $mathbb{R}$, but not paracompact.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Same question as the one I asked Captain Lama!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:56










    • $begingroup$
      The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
      $endgroup$
      – Eric Wofsey
      Apr 21 '16 at 17:59










    • $begingroup$
      @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 18:04
















    1












    1








    1





    $begingroup$

    A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $mathbb{R}$, but not paracompact.






    share|cite|improve this answer









    $endgroup$



    A locally contractible space need not even be paracompact. For instance, the long line is locally homeomorphic to $mathbb{R}$, but not paracompact.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 21 '16 at 17:54









    Eric WofseyEric Wofsey

    184k13212338




    184k13212338












    • $begingroup$
      Same question as the one I asked Captain Lama!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:56










    • $begingroup$
      The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
      $endgroup$
      – Eric Wofsey
      Apr 21 '16 at 17:59










    • $begingroup$
      @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 18:04




















    • $begingroup$
      Same question as the one I asked Captain Lama!
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 17:56










    • $begingroup$
      The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
      $endgroup$
      – Eric Wofsey
      Apr 21 '16 at 17:59










    • $begingroup$
      @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
      $endgroup$
      – Patrick Da Silva
      Apr 21 '16 at 18:04


















    $begingroup$
    Same question as the one I asked Captain Lama!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:56




    $begingroup$
    Same question as the one I asked Captain Lama!
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 17:56












    $begingroup$
    The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
    $endgroup$
    – Eric Wofsey
    Apr 21 '16 at 17:59




    $begingroup$
    The sheafification map of a presheaf need not be surjective on all open sets even if $X$ is hereditarily paracompact and locally contractible. For instance, if $X={0,1}$, then there is no reason for $mathcal{F}(X)tomathcal{F}^+(X)=mathcal{F}({0})timesmathcal{F}({1})$ to be surjective.
    $endgroup$
    – Eric Wofsey
    Apr 21 '16 at 17:59












    $begingroup$
    @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 18:04






    $begingroup$
    @Sorry, there was some axiom assumed on the sheaf. I think everything should be well-edited now.
    $endgroup$
    – Patrick Da Silva
    Apr 21 '16 at 18:04













    1












    $begingroup$

    To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility,
    in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3),
    where he also explains how Ramanan uses the hereditary paracompactness condition.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility,
      in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3),
      where he also explains how Ramanan uses the hereditary paracompactness condition.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility,
        in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3),
        where he also explains how Ramanan uses the hereditary paracompactness condition.






        share|cite|improve this answer









        $endgroup$



        To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility,
        in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3),
        where he also explains how Ramanan uses the hereditary paracompactness condition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 23 '16 at 18:50









        Dmitri PavlovDmitri Pavlov

        55326




        55326






























            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%2f1753095%2fare-locally-contractible-spaces-hereditarily-paracompact%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