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I am a little confused about what exactly are the difference(s) between simplicial complex, $\Delta$-complex, and CW Complex.

What I roughly understand is that $\Delta$-complexes are generalisation of simplicial complexes (without the requirement that the intersection of two simplicial complexes is another simplicial complex), and CW Complex further generalises that (how?).

Any explanation will be greatly appreciated.

Thanks for any enlightenment!

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Simplicial complexes, $\Delta$-complexes, and CW-complexes are all constructed by gluing together simplices. However, for each one, there are different rules for what kinds of "gluings" you are allowed to use.

For CW-complexes, you are allowed to use almost any gluing. Specifically, a CW-complex is constructed by induction, where at each step, you adjoin a new simplex by gluing its boundary to the complex you have already by any map. More explicitly, if $Y$ is the CW-complex you have built so far and $f:\partial \Delta^n\to Y$ is any continuous map, you can build a CW-complex $X=Y\sqcup\Delta^n/{\sim}$, where $\sim$ is the equivalence relation that identifies any $x\in\partial\Delta^n$ with $f(x)\in Y$. The only restriction to this gluing process is that you have to add simplices in increasing order of dimension. That is, you have to start with all the $0$-simplices, then glue in all the $1$-simplices, then glue in all the $2$-simplices, and so on. You're not allowed to glue in a new $1$-simplex once you've already added a $2$-simplex. (If you drop this ordering condition, you get a more general notion, which is sometimes called simply a "cell complex".)

For $\Delta$-complexes, you do the same thing, except that the maps $f$ you can use when adding a new cell are highly restricted. Specifically, for each $(n-1)$-simplex $A$ which is a face of $\partial\Delta^n$, the restriction of $f$ to $A$ must be equal to the inclusion of one of the $(n-1)$-simplices you already have. That is, $f$ maps the $n$ vertices of $A$ (with their canonical ordering) to the $n$ vertices of some $(n-1)$-simplex you've already added to your complex (with the same ordering on the vertices), and $f$ extends to all of $A$ by just interpolating linearly. Intuitively, this means that your complex is a union of simplices which are glued together by just gluing their faces together in the "obvious" linear way (for instance, as one encounters in a triangulation of a surface), rather than by arbitrary complicated continuous maps. Note, however, that some faces of a single simplex might get glued to each other: the restriction on what $f$ can be only applies to each face of $\partial\Delta^n$ separately. So, for instance, you can start with a single vertex, and then add an edge both of whose boundary vertices are the one vertex you started with (this gives you a circle). You could then add a triangle such that each of its three sides are equal to the one edge you have (this gives a space which cannot be embedded in $\mathbb{R}^3$ and is rather hard to visualize!).

Finally, simplicial complexes are $\Delta$-complexes which satisfy even more restrictions. First, $f$ is required to map different faces of $\partial\Delta^n$ to different $(n-1)$-simplices, so the situation discussed at the end of the previous paragraph cannot happen. In addition, you are not allowed to add two different $n$-simplices with the same set of vertices, so that a simplex in a simplicial complex is uniquely determined by its set of vertices (which, by the first requirement, you can show are all distinct).

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  • $\begingroup$ Thanks, very enlightening! Just to check, this diagram of a circle with one edge and one vertex is a delta-complex? math.stackexchange.com/q/1527329/28012 $\endgroup$ – yoyostein Nov 14 '15 at 5:13
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    $\begingroup$ Yes, it is a $\Delta$-complex (but not a simplicial complex). $\endgroup$ – Eric Wofsey Nov 14 '15 at 5:51
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    $\begingroup$ Hi Eric, my understanding of the definition of a simplicial complex is that it is a collection of simplices in $\mathbb{R}^{n}$ where the faces of each of these simplices is part of the collection and the intersection of any two simplices is a face of each of them. How does the example of the circle you gave fail this? $\endgroup$ – user135520 Jan 18 '16 at 20:25
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    $\begingroup$ @user135520: Are your simplices required to be linear (i.e., each simplex is actually the convex hull of its vertices)? If so, you can't draw a line segment in $\mathbb{R}^n$ whose two endpoints are equal... $\endgroup$ – Eric Wofsey Jan 18 '16 at 22:04

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