# What are the subobjects of a manifold?

Categorically a subobject of an object $a$ of some category $A$ is an object $a'$ with a monic morphism to $a$, ie $a'\to a$, upto isomorphism.

When $A$ is either a Topological or Differential manifold, what are the subobjects of a manifold, and are they the same as submanifolds?

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That depends. What morphisms are you choosing? –  Qiaochu Yuan Aug 23 '12 at 2:21
@Yuan: The real intent of the question is how to categorically select the 'usual' submanifolds, ie embeddings, immersed submanifolds etc which from peoples comments seem to require Diff being concrete; and to see whether the usual categorical subobjects result in anything useful here. –  Mozibur Ullah Aug 23 '12 at 21:10

Monic morphisms are injections in concrete categories like the category of manifolds, so a subobject of a manifold $M$ is another manifold with an injection into $M$. This isn't quite the same thing as a submanifold, as usually submanifolds are required to be embedded, meaning that they inherit their topology from the larger manifold.

For an example of a manifold that is injected into a larger manifold but isn't embedded, let $M$ be a torus (considered as a quotient space of $\mathbb{R}^2$ under integer translations), and let $L$ be the real line, mapped to a line of irrational slope in $\mathbb{R}^2$ and then projected to the torus. This causes the line to "wrap around" infinitely without touching itself. This map is an injection, thus $L$ with this map is a subobject, but the image isn't a manifold, since any neighborhood of a point contains infinitely many "nearby" lines.

Edit: As pointed out below, I was erroneous when I said monic morphisms are injective in all concrete categories. It is true for manifolds, however, as proved in Makoto Kato's answer.

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Ok, so to get an embedded manifold, we must stipulate that it has the initial topology wrt to the injection. That works for topological manifolds, but not for differential manifolds. –  Mozibur Ullah Aug 23 '12 at 0:05
"Monic morphisms are injections in concrete categories" This is not true in general. –  Makoto Kato Aug 23 '12 at 0:46
A sufficient condition for a concrete category $C$ to have the property that the monics are precisely the injective morphisms is that the forgetful functor $C \to \text{Set}$ is representable. This is true for the category of manifolds and for many other common concrete categories, but is false in general. In general we can only conclude that injections are monic. –  Qiaochu Yuan Aug 23 '12 at 2:18
@MakotoKato Thanks for pointing out my mistake, I've noted it in the bottom of the answer. To MoziburUllah: Yes, in the differentiable category you also require that the map is an immersion. I can't actually come up with with a smooth injective map of manifolds that isn't an immersion off the top of my head, though. –  MartianInvader Aug 23 '12 at 18:16
@MartianInvader As for an example of a smooth injective map of manifolds that is not an immersion, please see my answer. –  Makoto Kato Aug 23 '12 at 18:33

Let $\mathcal{C}$ be the category of toplogical(resp. differential) manifolds. The objects of $\mathcal{C}$ are topological(resp. differential) manifolds and the morphisms of $\mathcal{C}$ are continuous(resp. smooth) maps.

Let $f\colon X \rightarrow Y$ be a morphism in $\mathcal{C}$. We claim $f$ is a monomorphism if and only if $f$ is injective.

Suppose $f$ is a monomorphism. Let $x, y$ be distinct points of $X$. Let $p$ be a $0$-dimensional object in $\mathcal{C}$. There exists the unique morphism $g\colon p \rightarrow X$ such that $g(p) = x$. Similarly there exists the unique morphism $h\colon p \rightarrow X$ such that $h(p) = y$. Since $g \neq h$, $fg \neq fh$. Hence $f(x) \neq f(y)$. Hence $f$ is an injective map.

Conversly suppose $f$ is injective. Clearly $f$ is a monomorphism.

"Are they the same as submanifolds?"

Generally no.

Counter-example:

Let $f\colon \mathbb{R} \rightarrow \mathbb{R}^2$ be the map defined by $f(x) = (x^3, 0)$. $f$ is smooth and injective, but is not an immersion($f'(0) = 0$). Hence $\mathbb{R}$ cannot be identified with a submanifold of $\mathbb{R}^2$ by $f$.

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+1 Well explained! Perhaps it might be helpful for some less experienced readers if you would specify the objects and morphisms in the categories of topological/differential manifolds. –  magma Aug 23 '12 at 12:55
I always learned that a map isn't considered smooth if the derivative is ever zero. Otherwise, you could make a path with corners and call it "smooth" by having all its derivatives be zero at each corner. –  MartianInvader Aug 23 '12 at 18:37
@MartianInvader If you don't allow a smooth function to have the zero derivative, you can't call a constant function smooth. In your case, a path with corners is not differentiable at the corners. Regards, –  Makoto Kato Aug 23 '12 at 19:09
@Kato: From what I understand, an immersion implies locally the image is a submanifold. From this it doesn't follow that if it isn't an immersion then the image cannot be locally a submanifold. It seems to me it could go either way? –  Mozibur Ullah Aug 23 '12 at 21:29
@MoziburUllah An injective immersion is identified with the canonical injection of a submanifold. Conversely the canonical injection of a submanifold is an immersion. Regards, –  Makoto Kato Aug 23 '12 at 21:55