well, that's what a good multivariable calc course should start you on ... and then diff geo will advance you a lot if it's a good class (with pictures) :)
no, I don't think there is any retract $X \to A$. retract induces an injective map on $\pi_1$, and $\pi_1(S^1 \times I/~) \cong \Bbb Z*\Bbb Z$ doesn't inject into $\Bbb Z \times \Bbb Z$
If I compose two functions $f$ and $1-g$, do I have to put the $1-g$ in parentheses or is it sufficient to write $(f \circ 1-g)(x)$ (vs. $(f \circ (1-g))(x)$)?
well, the retract $T - \{pt\} \to S^1 \vee S^1$ is very explicit : take the fundamental square minus a point in the interior. now push points in the nbhds of the removed point to the boundary of the square, and identify afterwards. let me ponder on what is the preimage of an nbhd of the bad point of $S^1 \vee S^1$.
@TedShifrin i have no idea what you mean by description of the map, but the preimage of little nbhds of the wedged point is disjoint union of four small nbhds containing points close to the removed point of the torus.
oh, i see. well, that depends on the disk. in general, you push your disk to the boundary square of the fundamental square via sliding through a ray issuing from the removed point.
let us agree that we know of a continuous retract $f : D^2 - \{pt\} \to S^1$ (once, again i am not writing a formula). $g : S^1 \to S^1 \vee S^1$ be the quotient map obtained by identifying a square via the word $abab$. This is continuous. Composition of continuous maps is continuous. That said, I don't know why I should care about explicitly writing out the nbhds.
One of my old math friends decided to post my syllabus and problem set from a complex geometry course in 1980 on FB ... I asked him if it was part of my retirement roast. :)
@Ted Well, draw the lines y = x and y = -x in R^2. This partitions R^2 into 4 parts. Push each point in the interior of each of the single parts into the axis contained in these. Then push the points on the diagonal lines to the origin.
gah, i got disconnected and he's gone.
oh, and if (x, y) is a point in one of the conical parts, push it to the axis via pushing it through straightlines, which i forgot to add.
PS, @Mike, I saw someone ask in the comments in a recent question whether $\Bbb RP^2$ is a retract of the klein bottle. I really don't know how to prove it. What if there's some whacky embedding of $\Bbb RP^2$ inside the bottle to which the bottle retracts to?
@MikeMiller oh, I guess I have it. let $f$ be an embedding of a compact $n$-manifold $M$ into a connected compact $n$-manifold $N$. open cover $M$ with finitely many open sets homeomorphic to $\Bbb R^n$. embeddings take open sets to open sets, so $f(M) \subset N$ is open. but $M$ is compact, so it's also closed. impossible.
@BalarkaSen: The point is there's absolutely no hope of doing so in general. The best case for what you're trying to do is that $X$ is path-connected, and thus we have an extension of that form.
i'd love it if you stop going off-the-tangent while you're discussing stuff. :P when you mentioned the extension problem, i was like "huh?" although i knew what it is.
come to think of it, i think prof told me that there is provably no algorithm (<-- how to make this precise?) to compute fundamental group of a complex projective variety.
this came up as a reply to my retort about homology being way too algebraic and hard to compute when i was studying it, and that i thought fundamental group is much easier to compute.
moreover, he told me that there is such an algorithm for computing singular homology of complex projective varieties. i believe it now, the axioms provide a very strong way of computing homology of stuff
reminds me of something my prof recently said: there is no algorithm for proving that some two 2-dimensional cell complexes are homotopy equivalent. and this follows from the fact that for every group G there is a 2-dim cell complex whose fundamental group is G, and the "isomorphism problem"
@iwriteonbananas ah, yes. you'll do those when you do Cayley complexes. but is there a proof that there is no algorithm for solving the isomorphism problem? i don't know.
@BalarkaSen: I would phrase it a little differently. You can build a 4-manifold algorithmically given a presentation of a group. Solving the homeomorphism problem would, in particular, then tell you when two presentations are equal. It's not enough that you can build something given a group, it has to be from a presentation. i.e. you can classify $K(G,1)$s up to homotopy equivalence :P
To take this somewhere where this doesn't sound like pedantry: maybe this is why the classification of projective varieties up to homotopy equivalence isn't thought to be impossible? That there's no way to start with a presentation of a certain type and build a complex projective variety out of it.
It's of course not known that you can get every finitely presented group out of complex projective varieties (indeed, the opposite is true), but if you could do that construction for some sufficiently bad subset of all finitely presented groups where the isomorphism problem still doesn't work, that would cause problems.
So maybe it's a good thing for the classification that there's no good algorithm to calculate the fundamental group.
@iwriteonbananas they're nothing to bother about, but they are a good way to construct universal covers. here : let $G$ be a finitely presented group. let $\vee S^1$ be a bouquet of $n$ circles, each labelled with the generator of $G$. paste a cell to the bouquet according to each word in the presentation of $G$. voila.
@TedShifrin well, preimage of an open interval around a point in one of the axes away from origin is an open strip. preimage of an open interval around the origin is messes up, yikes.
