Exact additive functor preserves homology












0












$begingroup$



Let $R$ and $A$ be rings, $T:{}_{R}text{Mod}rightarrow {}_Atext{Mod}$ be an exact additive functor and $(C_bullet,d_bullet)$ a chain complex in ${}_{R}text{Mod}$. Prove that
$$H_n(TC_bullet,Td_bullet)cong TH_n(C_bullet,d_bullet) $$




I know this looks too easy, but I can't do an actual proof of this fact.



The thing is that is easy to prove that $Tker(d_n)cong ker(T d_n)$ and $T text{im}(d_{n+1})cong text{im}(Td_{n+1})$ but this isomorphims are to abstract so I can't prove that the latter isomorphism is a restriction of the former.



Also, I have the feeling that the additive hypothesis is not needed. Is that correct?






category-theory homology-cohomology homological-algebra abelian-categories







share|cite|improve this question











$endgroup$












  • $begingroup$
    I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
    $endgroup$
    – Walter Simon
    Oct 11 '17 at 0:10












  • $begingroup$
    you use additivity in the fact that $T$ maps chain complexes to chain complexes
    $endgroup$
    – M. Van
    Oct 11 '17 at 1:19






  • 1




    $begingroup$
    $T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
    $endgroup$
    – Hurkyl
    Oct 11 '17 at 16:18






  • 1




    $begingroup$
    @WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
    $endgroup$
    – M. Van
    Oct 11 '17 at 18:35






  • 2




    $begingroup$
    @M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
    $endgroup$
    – Roland
    Oct 11 '17 at 20:25
















0












$begingroup$



Let $R$ and $A$ be rings, $T:{}_{R}text{Mod}rightarrow {}_Atext{Mod}$ be an exact additive functor and $(C_bullet,d_bullet)$ a chain complex in ${}_{R}text{Mod}$. Prove that
$$H_n(TC_bullet,Td_bullet)cong TH_n(C_bullet,d_bullet) $$




I know this looks too easy, but I can't do an actual proof of this fact.



The thing is that is easy to prove that $Tker(d_n)cong ker(T d_n)$ and $T text{im}(d_{n+1})cong text{im}(Td_{n+1})$ but this isomorphims are to abstract so I can't prove that the latter isomorphism is a restriction of the former.



Also, I have the feeling that the additive hypothesis is not needed. Is that correct?






category-theory homology-cohomology homological-algebra abelian-categories







share|cite|improve this question











$endgroup$












  • $begingroup$
    I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
    $endgroup$
    – Walter Simon
    Oct 11 '17 at 0:10












  • $begingroup$
    you use additivity in the fact that $T$ maps chain complexes to chain complexes
    $endgroup$
    – M. Van
    Oct 11 '17 at 1:19






  • 1




    $begingroup$
    $T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
    $endgroup$
    – Hurkyl
    Oct 11 '17 at 16:18






  • 1




    $begingroup$
    @WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
    $endgroup$
    – M. Van
    Oct 11 '17 at 18:35






  • 2




    $begingroup$
    @M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
    $endgroup$
    – Roland
    Oct 11 '17 at 20:25














0












0








0





$begingroup$



Let $R$ and $A$ be rings, $T:{}_{R}text{Mod}rightarrow {}_Atext{Mod}$ be an exact additive functor and $(C_bullet,d_bullet)$ a chain complex in ${}_{R}text{Mod}$. Prove that
$$H_n(TC_bullet,Td_bullet)cong TH_n(C_bullet,d_bullet) $$




I know this looks too easy, but I can't do an actual proof of this fact.



The thing is that is easy to prove that $Tker(d_n)cong ker(T d_n)$ and $T text{im}(d_{n+1})cong text{im}(Td_{n+1})$ but this isomorphims are to abstract so I can't prove that the latter isomorphism is a restriction of the former.



Also, I have the feeling that the additive hypothesis is not needed. Is that correct?






category-theory homology-cohomology homological-algebra abelian-categories







share|cite|improve this question











$endgroup$





Let $R$ and $A$ be rings, $T:{}_{R}text{Mod}rightarrow {}_Atext{Mod}$ be an exact additive functor and $(C_bullet,d_bullet)$ a chain complex in ${}_{R}text{Mod}$. Prove that
$$H_n(TC_bullet,Td_bullet)cong TH_n(C_bullet,d_bullet) $$




I know this looks too easy, but I can't do an actual proof of this fact.



The thing is that is easy to prove that $Tker(d_n)cong ker(T d_n)$ and $T text{im}(d_{n+1})cong text{im}(Td_{n+1})$ but this isomorphims are to abstract so I can't prove that the latter isomorphism is a restriction of the former.



Also, I have the feeling that the additive hypothesis is not needed. Is that correct?






category-theory homology-cohomology homological-algebra abelian-categories




category-theory homology-cohomology homological-algebra abelian-categories






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 11 '17 at 1:14







Walter Simon

















asked Oct 10 '17 at 23:59









Walter SimonWalter Simon

15410




15410












  • $begingroup$
    I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
    $endgroup$
    – Walter Simon
    Oct 11 '17 at 0:10












  • $begingroup$
    you use additivity in the fact that $T$ maps chain complexes to chain complexes
    $endgroup$
    – M. Van
    Oct 11 '17 at 1:19






  • 1




    $begingroup$
    $T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
    $endgroup$
    – Hurkyl
    Oct 11 '17 at 16:18






  • 1




    $begingroup$
    @WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
    $endgroup$
    – M. Van
    Oct 11 '17 at 18:35






  • 2




    $begingroup$
    @M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
    $endgroup$
    – Roland
    Oct 11 '17 at 20:25


















  • $begingroup$
    I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
    $endgroup$
    – Walter Simon
    Oct 11 '17 at 0:10












  • $begingroup$
    you use additivity in the fact that $T$ maps chain complexes to chain complexes
    $endgroup$
    – M. Van
    Oct 11 '17 at 1:19






  • 1




    $begingroup$
    $T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
    $endgroup$
    – Hurkyl
    Oct 11 '17 at 16:18






  • 1




    $begingroup$
    @WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
    $endgroup$
    – M. Van
    Oct 11 '17 at 18:35






  • 2




    $begingroup$
    @M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
    $endgroup$
    – Roland
    Oct 11 '17 at 20:25
















$begingroup$
I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
$endgroup$
– Walter Simon
Oct 11 '17 at 0:10






$begingroup$
I know there is already a question in the site about this. But the solution given there have some problems. For example the equality $F(partial_{n+1})(F(C_{n+1})))=F(partial_{n+1}(C_{n+1}))$ is not true without exactness an even with exactness it is just an isomorphism. So you need to show that this isomorphism relates well with respect to the other isomorphism of the kernel. Otherwise is not true the part $$frac{ker(F(partial_n))}{Im(F(partial_{n+1}))}stackrel{by (1)(2)}{cong}frac{F(ker(partial_n))}{F(Im(partial_{n+1}))}$$.
$endgroup$
– Walter Simon
Oct 11 '17 at 0:10














$begingroup$
you use additivity in the fact that $T$ maps chain complexes to chain complexes
$endgroup$
– M. Van
Oct 11 '17 at 1:19




$begingroup$
you use additivity in the fact that $T$ maps chain complexes to chain complexes
$endgroup$
– M. Van
Oct 11 '17 at 1:19




1




1




$begingroup$
$T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
$endgroup$
– Hurkyl
Oct 11 '17 at 16:18




$begingroup$
$T ker(d_n)$ is a kernel of $T d_n$. Your issue boils down to how to understand in what way homology is independent of which specific choices of kernels, images, and quotients you use in its construction.
$endgroup$
– Hurkyl
Oct 11 '17 at 16:18




1




1




$begingroup$
@WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
$endgroup$
– M. Van
Oct 11 '17 at 18:35




$begingroup$
@WalterSimon we need $Td_i circ Td_{i-1}=0$, for this we need $T$ to map the $0$ map to the $0$ map and again for this we need additivity. But maybe it follows from exactness as well (?)
$endgroup$
– M. Van
Oct 11 '17 at 18:35




2




2




$begingroup$
@M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
$endgroup$
– Roland
Oct 11 '17 at 20:25




$begingroup$
@M.Van Yes this follows from exactness. Indeed, an exact functor preserve the zero object, and thus preserve zero maps (these are the maps which factors through the zero object). In fact, an exact functor also preserve direct sums and therefore it will necessarily be additive.
$endgroup$
– Roland
Oct 11 '17 at 20:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

Thanks to the comments I got with a proof that convince me.



As $T$ is left exact is easy to show that $ker (Td_n)cong Tker (d_n)$ as subobjects of $C_n$. Similarly using both right and left exactness one can prove that $text{im}(Td_{n+1})cong T text{im}(d_{n+1})$ as subobjects of $C_n$.



Therefore we have the diagram



diagram



Where $B_n=text{im}(d_{n+1})$, $B_n^T=text{im}(Td_{n+1})$, and so on



Now using the commutative triangles (given by the isomorphism at the level of subobjects) we can prove that $$(TB_nrightarrow TZ_nrightarrow Z_n^Trightarrow C_n)=(TB_nrightarrow B_n^Trightarrow Z_n^Trightarrow C_n)$$



And then as $Z_n^Trightarrow C_n$ is a monomorphism the square is commutative. Then we can finish the proof constructing $TH_ncong H_n^T$ by a diagram chasing or using the universal property of the cokernel.






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%2f2466829%2fexact-additive-functor-preserves-homology%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Thanks to the comments I got with a proof that convince me.



    As $T$ is left exact is easy to show that $ker (Td_n)cong Tker (d_n)$ as subobjects of $C_n$. Similarly using both right and left exactness one can prove that $text{im}(Td_{n+1})cong T text{im}(d_{n+1})$ as subobjects of $C_n$.



    Therefore we have the diagram



    diagram



    Where $B_n=text{im}(d_{n+1})$, $B_n^T=text{im}(Td_{n+1})$, and so on



    Now using the commutative triangles (given by the isomorphism at the level of subobjects) we can prove that $$(TB_nrightarrow TZ_nrightarrow Z_n^Trightarrow C_n)=(TB_nrightarrow B_n^Trightarrow Z_n^Trightarrow C_n)$$



    And then as $Z_n^Trightarrow C_n$ is a monomorphism the square is commutative. Then we can finish the proof constructing $TH_ncong H_n^T$ by a diagram chasing or using the universal property of the cokernel.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Thanks to the comments I got with a proof that convince me.



      As $T$ is left exact is easy to show that $ker (Td_n)cong Tker (d_n)$ as subobjects of $C_n$. Similarly using both right and left exactness one can prove that $text{im}(Td_{n+1})cong T text{im}(d_{n+1})$ as subobjects of $C_n$.



      Therefore we have the diagram



      diagram



      Where $B_n=text{im}(d_{n+1})$, $B_n^T=text{im}(Td_{n+1})$, and so on



      Now using the commutative triangles (given by the isomorphism at the level of subobjects) we can prove that $$(TB_nrightarrow TZ_nrightarrow Z_n^Trightarrow C_n)=(TB_nrightarrow B_n^Trightarrow Z_n^Trightarrow C_n)$$



      And then as $Z_n^Trightarrow C_n$ is a monomorphism the square is commutative. Then we can finish the proof constructing $TH_ncong H_n^T$ by a diagram chasing or using the universal property of the cokernel.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Thanks to the comments I got with a proof that convince me.



        As $T$ is left exact is easy to show that $ker (Td_n)cong Tker (d_n)$ as subobjects of $C_n$. Similarly using both right and left exactness one can prove that $text{im}(Td_{n+1})cong T text{im}(d_{n+1})$ as subobjects of $C_n$.



        Therefore we have the diagram



        diagram



        Where $B_n=text{im}(d_{n+1})$, $B_n^T=text{im}(Td_{n+1})$, and so on



        Now using the commutative triangles (given by the isomorphism at the level of subobjects) we can prove that $$(TB_nrightarrow TZ_nrightarrow Z_n^Trightarrow C_n)=(TB_nrightarrow B_n^Trightarrow Z_n^Trightarrow C_n)$$



        And then as $Z_n^Trightarrow C_n$ is a monomorphism the square is commutative. Then we can finish the proof constructing $TH_ncong H_n^T$ by a diagram chasing or using the universal property of the cokernel.






        share|cite|improve this answer











        $endgroup$



        Thanks to the comments I got with a proof that convince me.



        As $T$ is left exact is easy to show that $ker (Td_n)cong Tker (d_n)$ as subobjects of $C_n$. Similarly using both right and left exactness one can prove that $text{im}(Td_{n+1})cong T text{im}(d_{n+1})$ as subobjects of $C_n$.



        Therefore we have the diagram



        diagram



        Where $B_n=text{im}(d_{n+1})$, $B_n^T=text{im}(Td_{n+1})$, and so on



        Now using the commutative triangles (given by the isomorphism at the level of subobjects) we can prove that $$(TB_nrightarrow TZ_nrightarrow Z_n^Trightarrow C_n)=(TB_nrightarrow B_n^Trightarrow Z_n^Trightarrow C_n)$$



        And then as $Z_n^Trightarrow C_n$ is a monomorphism the square is commutative. Then we can finish the proof constructing $TH_ncong H_n^T$ by a diagram chasing or using the universal property of the cokernel.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 10:58

























        answered Oct 11 '17 at 18:20









        Walter SimonWalter Simon

        15410




        15410






























            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%2f2466829%2fexact-additive-functor-preserves-homology%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