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The study of differentiable manifolds and differentiable maps. One fundamental problem is that of classifying manifolds up to diffeomorphism. Differential topology is what Poincaré understood as topology or “analysis situs”.

3 votes
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a small questions about hopf theorem

I realize that this doesn't answer your question, but there is also an approach using the methods of homotopy theory and CW complexes. If $M$ is a closed smooth orientable $p$-manifold, then $M$ is ho …
Ian Agol's user avatar
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9 votes

On the generalized Gauss-Bonnet theorem

I'll make some general remarks. One may break down the generalized Gauss-Bonnet into two parts. The first is to verify that it is a topological invariant. This means that it should be independent of t …
Ian Agol's user avatar
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10 votes

Isotopy in 3-manifolds

If $\Sigma_1 \hookrightarrow M$ is an embedded $\pi_1$-injective surface, then any homotopic embedded surface will be isotopic to $\Sigma$. As Ryan and Allen point out, this is due to Waldhausen for i …
Ian Agol's user avatar
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43 votes

Is differential topology a dying field?

I don't think differential topology is a dying field. I'll interpret this as the classification of smooth manifolds and, more broadly, maps between them (immersions, embeddings, diffeomorphism group …
Ian Agol's user avatar
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6 votes
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Irreducibility of 3-manifolds with (non)empty boundary

You're asking how reducibility/irreducibility behaves under drilling and filling. I think you've captured the essence of drilling: if a link is "sphere busting" in a reducible manifold (meets every es …
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5 votes
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what is the meaning of "inseparable" in this case

I think you mean "non-separating" rather than "inseparable" (or at least I'm not familiar with that terminology), which means as you say that when you cut along the submanifold, you get a connected sp …
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9 votes
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General position for map from surface to 3-manifold

In general, no you cannot. Consider one dimension down. Take two curves on a torus, intersecting transversely in a point. One of the curves cannot be homotoped to be disjoint from the pair of curves. …
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9 votes
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When can a surface in a 3-manifold be isotoped off a knot?

Let’s assume that the manifold $M$ is irreducible and orientable and the surface $S$ is orientable. This is to avoid 1-sided surfaces. First let’s assume that the surface $S$ is fully compressible. T …
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4 votes
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Compact closed aspherical manifolds with vanishing second homology in all the covering spaces

I think that the answer to this question is unknown in general. If one had a closed aspherical manifold with this property, then it could not contain a Baumslag-Solitar subgroup since such a group has …
Ian Agol's user avatar
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5 votes

Classification of surface bundles over surfaces

For a), when $\chi(F)<0$ $F$-bundles over $B$ are classified by maps $\pi_1(B)\to Mod(F)$, where $Mod(F)$ is the mapping class group of $F$. This boils down to the fact that $Diff_0(F)$ is contractibl …
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7 votes

Embedded 2-tori in $S^1\times S^4$

I think this is true (homotopy implies isotopy). Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a circle-valued Morse function on $T^2$. Becau …
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8 votes
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Realizing Morse functions on $S^2$ as height functions

Any Morse function on $S^2$ may be realized by an embedding $S^2\hookrightarrow \mathbb{R}^3 \to \mathbb{R}$. For a Morse function $F:S^2\to \mathbb{R}$, take the equivalence relation with equivalence …
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7 votes
Accepted

Smooth map between oriented manifolds

This follows from a result of Hopf (see the exposition of Epstein). By this result, one may assume that there is a disk $D\subset N$ such that $f^{-1}(D)$ is $d$ disks mapping diffeomorphically to $D$ …
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5 votes

Homologically trivial submanifolds

This is true in codimension and dimension 1. In fact, in these dimensions, any homology class is represented by an embedded manifold. In codimension 1, the manifold $N$ will separate $M$ into (at lea …
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5 votes

Theorems similar to Tischler fibering theorem

On a $2k+1$-dimensional manifold, a 1-form $\alpha$ such that $\alpha\wedge (d\alpha)^k \neq 0$ at each point gives a contact structure. I believe that it is still open which $2k+1$-manifolds for $k>1 …
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