Final step of homotopy lemma

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In proving Homotopy lemma in Milnor Topology from the differential viewpoint we consider $V_1 cap V_2$ where $V_1$ is a neighbourhood of $y$ in which $card f^-1( y)$ is constant. Similarly $V_2$ s a neighbourhood of $y$ in which $card g^-1( y)$ is constant. If $F$ is smooth homotopy between $f$ and $g$ then he chooses $zin V_1cap V_2$ such that $z$ is regular value of $F$. How do we know such $z$ exists ?

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    In proving Homotopy lemma in Milnor Topology from the differential viewpoint we consider $V_1 cap V_2$ where $V_1$ is a neighbourhood of $y$ in which $card f^-1( y)$ is constant. Similarly $V_2$ s a neighbourhood of $y$ in which $card g^-1( y)$ is constant. If $F$ is smooth homotopy between $f$ and $g$ then he chooses $zin V_1cap V_2$ such that $z$ is regular value of $F$. How do we know such $z$ exists ?

    Let me know if more information is needed.







    share|cite|improve this question





















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      In proving Homotopy lemma in Milnor Topology from the differential viewpoint we consider $V_1 cap V_2$ where $V_1$ is a neighbourhood of $y$ in which $card f^-1( y)$ is constant. Similarly $V_2$ s a neighbourhood of $y$ in which $card g^-1( y)$ is constant. If $F$ is smooth homotopy between $f$ and $g$ then he chooses $zin V_1cap V_2$ such that $z$ is regular value of $F$. How do we know such $z$ exists ?

      Let me know if more information is needed.







      share|cite|improve this question











      In proving Homotopy lemma in Milnor Topology from the differential viewpoint we consider $V_1 cap V_2$ where $V_1$ is a neighbourhood of $y$ in which $card f^-1( y)$ is constant. Similarly $V_2$ s a neighbourhood of $y$ in which $card g^-1( y)$ is constant. If $F$ is smooth homotopy between $f$ and $g$ then he chooses $zin V_1cap V_2$ such that $z$ is regular value of $F$. How do we know such $z$ exists ?

      Let me know if more information is needed.









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      asked Jul 16 at 13:53









      mathemather

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          $V_1 cap V_2$ is non-empty (since $y$ is in both terms) and open. If there were no regular values for $F$ there, you'd have a set of positive measure containing only critical values. This contradicts Sard. (See also Brown's corollary on page 11 of the Princeton Landmarks version.)






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            $V_1 cap V_2$ is non-empty (since $y$ is in both terms) and open. If there were no regular values for $F$ there, you'd have a set of positive measure containing only critical values. This contradicts Sard. (See also Brown's corollary on page 11 of the Princeton Landmarks version.)






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              $V_1 cap V_2$ is non-empty (since $y$ is in both terms) and open. If there were no regular values for $F$ there, you'd have a set of positive measure containing only critical values. This contradicts Sard. (See also Brown's corollary on page 11 of the Princeton Landmarks version.)






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                $V_1 cap V_2$ is non-empty (since $y$ is in both terms) and open. If there were no regular values for $F$ there, you'd have a set of positive measure containing only critical values. This contradicts Sard. (See also Brown's corollary on page 11 of the Princeton Landmarks version.)






                share|cite|improve this answer















                $V_1 cap V_2$ is non-empty (since $y$ is in both terms) and open. If there were no regular values for $F$ there, you'd have a set of positive measure containing only critical values. This contradicts Sard. (See also Brown's corollary on page 11 of the Princeton Landmarks version.)







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Jul 16 at 14:14


























                answered Jul 16 at 14:09









                Randall

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