Dunce Cap triangle homotopy equivalent to $S^1$ 
Why is Dunce Cap triangle homotopy equivalent to $S^1$. 

Yes, maybe it is not the correct expression, but i mean that the dunce cap can be represented as a quotient space of a triangle by identifying the 3 sides as pictured below (1. picture from left).
and if you consider only the edges, this should be homotopy equivalent to $S^1$, and i ask, if my ''transformation'' in the picture is correct.

 A: Your proof looks good and would be accepted by anyone reasonable marking it. (Perhaps throw in some words as well so that the reader is clear about what the image means.)
As far as a rigorous proof goes. I would start by showing that the space, let us call it $D$, is compact (it's the continuous image of a compact space so is compact) and that $D$ is a topological manifold. This can be seen by noting that each of the vertices are mapped to the same point in the quotient and a small neighbourhood of this point is homeomorphic to an open interval. The image of the points on the interior of the edges also clearly have a small neighbourhood homeomorphic to an open interval. It follows that $D$ is a compact connected topological $1$-dimensional manifold without boundary and so can only be a circle up to homeomorphism.
This proof is obviously overkill (and uses a fairly strong theorem - $1$-manifolds are disjoint unions of circles and real lines) and a picture really is the best way to go here. I am still slightly concerned why you mention the weaker result that $D$ is homotopy equivalent to $S^1$. Is it possible you really want to show that the 'punctured dunce cap' is homotopy equivalent to the circle?
A: The term Dunce Cap is usually applied to the adjunction space $S^1 \cup_f D^2$ where the disc $D^2$ is attached to the circle $S^1$ by a map which we can write $a +a -a$ But $$a +a -a\simeq a+(a-a)\simeq a+0 \simeq a,$$ so $f$ is homotopic to the identity map $i: S^1 \to S^1$. So the Dunce Cap as defined here is homotopy equivalent to $S^1 \cup_i D^2$, which is clearly homeomorphic to $D^2$, and so is contractible. See p. 295 of Topology and Groupoids. I think this is what you mean by your argument, which has caught the essential idea. 
You might like a picture (from the above book) of the identifications up to the last stage:

The next stage is to identify the two arrowed circles. I have made this in plasticine. 
A: I'm just looking at the dunce's cap topology for the first time.
Reading the wikipedia article it states:

The dunce hat is contractible, but not collapsible. Contractibility can be easily seen by noting that the dunce hat embeds
  in the 3-ball and the 3-ball deformation retracts onto the dunce hat.
  Alternatively, note that the dunce hat is the CW-complex obtained by
  gluing the boundary of a 2-cell onto the circle.

So if it is the case that it's contractible then I think that makes it homotopy equivalent to a point, again from wikipedia:

A contractible space is precisely one with the homotopy type of a point. It follows that all the homotopy groups of a contractible space
  are trivial.

I know I'm not offering any new proofs myself but thought this was an interesting answer.
