If I look at this, I get t If one's proofs are easier to read and manage to clearly show the method, one can often get away with inaccuracies, whereas they'll always be noted if one has to figure out the meaning of every word or symbol.he feeling that the main problem might actually be style. I'll at the moment only look at the first part and it is clear from it that you know what's going on. However, you seem to have a hard time conveying this to the person who has to read it (I guess some teaching assistant). I don't mean to be degrading or insulting and hope you can somehow use what I'm writing. An easy bit of advise would be to read your own proofs when you're done, as if you didn't know the solution yet. Would you still understand what happens?
More concretely, what would help the first part? First of all, if there are two 'kinds of maps' involved, just give them different names. When there are maps $X\to Y$ and $\to X$, which is the case right now, don't call all of them $f$ with some added decorations. Once you need lots of maps, you'll have to reuse the same letter sometimes, but you could easily add a letter in this case. Then, where does the first sentence ('First we show that...') come from? Contractibility of $X$! Just say so and the TA doesn't need to think about this anymore. Also, do you really just want $f_0(X)=X$ instead of $f_0=\text{id}_x$? (and again, later on, you want $f=\hat{f}_0$ - not just the same image)
This may well be an illustration of what I wrote before. It is an error and at the moment very likely to go noticed - even when your thoughts are clearly correct. Would you have written the following containing the same error, it might just pass a TA without him/her noticing: "Since the space $X$ is contractible, there exists a continuous family of maps $(f_t)$ with $t\in[0,1]$, such that $f_0(X)=X$ and $f_1(X)=x$.
I think the biggest inaccuracy is gone now. Just one small thing: once you define $\hat{f}$, you first lose the subscript $t$ and then all of a sudden it's there again. This should be there all the time. Also not wrong but making it harder for the reader: if $\mathbb{I}$ is a map from $X$ to $X$, you can just as well omit the $|_X$.
Let $\tilde a,\tilde b:I\rightarrow X$ be lifts of $a,b:I\rightarrow Y$ satisfying $p\circ \tilde a=a$, $p\circ \tilde b=b$, and let $F_t:a\simeq_{\partial I}b:I\rightarrow Y$ be a relative homotopy satisfying $F(s,0)=a(s)$, $F(s,1)=b(s)$, $F(0,t)=a(0)=b(0)$ and $F(1,t)=a(1)=b(1)$. Now apply the HELP lemma in the diagram
$\require{AMScd}$
\begin{CD}
0\times I\cup I\times \partial I@>G'>> X\\
@Vi V V @VVpV\\
I\times I @>F>> Y
\end{CD}$\require{AMScd}$
where $G'(s,0)=\tilde a(s)$, $G'(s,1)=\tilde b(s)$, $G'(0,t)=\tilde a(0)=\tilde b(0)$. Note that $p\circ G'=F\circ i$. From the data the HELP lemma gives a map $G:I\times I\rightarrow X$ satisfying
$$G(s,0)=\tilde a (s),\quad G(s,1)=\tilde b(s),\quad G(0,t)=\tilde a(0)=\tilde b(0),\quad p\circ G(s,t)=F(s,t).$$
In particular $p(G(s,1))=p(\tilde a(1))=p(\tilde b(1))=F(s,1)=a(0)=b(0)$, so the assignment $s\mapsto G(s,1)$ is a path in $X$ contained in the fibre over $a(0)=b(0)$, going from $\tilde a(0)$ and ending at $\tilde b(0)$. Since the covering projection has discrete fibres such a path must be constant.
Hence $\tilde a(1)=\tilde b(1)$, and $G:\tilde a\simeq_{\partial I}\tilde b$ is a relative homotopy.
Best Answer
That's a theorem, actually a really important one about covering spaces.
This theorem states that
note: the theorem deals with pointed spaces because of the uniqueness requirement if you forget about fixed points the the result implies the existance of a lifting but you lose the uniqueness requirement.
You can apply this theorem since $\pi_1(S^2) \cong \pi_1(\mathbb R)=0$ are the trivial group, so the image of the $f_*$ and $p_*$ are the trivial group. From that you can get a lift $\tilde f$.