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"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Even more is true: the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension two, in the sense that any smooth map $[0,1] \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed, relative to its endpoints $\lbrace 0,1\rbrace$, to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times [0,1]\to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.

"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Even more is true: the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension two, in the sense that any smooth map $D^1 \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed, relative to its boundary $S^0$, to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times D^1\to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.

"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Furthermore, the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension one, in the sense that any smooth map $I \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed relative to endpoints to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times I \to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.

"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Even more is true: the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension two, in the sense that any smooth map $[0,1] \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed, relative to its endpoints $\lbrace 0,1\rbrace$, to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times [0,1]\to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.

"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Then $C^\infty(M)$ is a contractible space (in fact it's affine). Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Furthermore, the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension one, in the sense that any smooth map $I \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed relative to endpoints to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times I \to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

"Generic" usually refers to open and dense.

Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)

Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Furthermore, the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension one, in the sense that any smooth map $I \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed relative to endpoints to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times I \to \Bbb R$ is a smooth map.

This is discussed in beautiful detail in the book:

http://www.amazon.com/Stable-Mappings-Singularities-Graduate-Mathematics/dp/0387900721/ref=la_B001HCW72E_1_4?ie=UTF8&qid=1366920811&sr=1-4

In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.