Space $W$ of null homotopic maps $g \in Hom(S^1, S^1)$ with $g(1)=1$ is contractible. This is the claim the proof of which I can't fully understand. In my question $S^1$ is defined as subset of complex numbers of unit magnitude.
The proof goes as follows:

We can obtain a bijection $V \to W$, where $V$ is a vector subspace (and thus contractible) of $Hom(S^1, \mathbb{R})$ of functions $f$ with $f(1)=0$. Bijection is obtained by composing with the function $p: \mathbb{R} \to S^1; x \mapsto \exp(2\pi i x)$. Function that composes with $p$ on left is denoted as $p_*$. We want to show that this bijection is open, that is, a homeomorphism. Since it is also a group homomorphism, it suffices to check that an open neighborhood of a neutral element is mapped to an open neighborhood of a neutral element. It is enough to consider sets of form $C(K, U)$, where $K \subseteq S^1$ is compact and $U \subseteq (-1, 1)$ is an open neighborhood of $0$. Since $p$ is open, then $p(U) \subseteq S^{-1} \setminus -1$ (why?) is open in $S^1$. It holds that $p_* C(K, U) = C(K, p(U))$ because of the initial conditions, and it is open.

$C(K, U)$ denotes a subbasic open set in compact-open topology. $Hom(S^1, \mathbb{R})$ is a group with pointwise addition, and $Hom(S^1, S^1)$ is a group with pointwise multiplication. Neutral elements in each are constant maps to $0$ and $1$ respectively. I can also prove $W$ is a subgroup of $Hom(S^1, S^1)$. $p_*$ is continuous because domains are locally compact.
Firstly, I don't understand why we only consider $U \subseteq (-1, 1)$. I also think that the claim that $p(U) \subseteq S^{-1} \setminus -1$ is false, because for it to be true we'd need to have $U \subseteq (-1/2, 1/2)$. But then we don't consider all possible neighborhoods, and so it doesn't prove that all open sets are mapped to open sets. I also don't get why we need the condition $p(U) \subseteq S^{-1} \setminus -1$ at all. Isn't it enough to just cite that $p$ is open so that $p_* C(K, U) = C(K, p(U))$ is open for an arbitrary $U$? And what could possibly initial conditions mean here?
Secondly, I'm not really sure why $p_*$ defines a bijection. I think I can argue that null homotopic maps can be lifted, which shows surjectivity, and also that lifts of the same function on differ by an integer constant, so there is only one of them with $f(1)=0$. Is it enough? By lifting a function $g: X \to S^1$ I mean finding a function $f: X\to \mathbb{R}$ such that $pf=g$. Two lifts differ by an integer constant because $pf = pf^{\prime}$ implies $pf/pf^{\prime}=1$, which says that $\exp(2\pi i(f - f^{\prime}))=1$.
I will now write a (special case of) Homotopy lifting property from my textbook that I use for my argument.

Let $X$ be a connected topological space, $f: X\to \mathbb{R}$ continuous function and $H: I \times X \to S^1$ a homotopy that starts with $pf$. Then there is a unique homotopy $F: I\times X \to \mathbb{R}$ that starts with $f$ and $pF = H$. 

I used it to say that null homotopic maps can be lifted.
Can it be used to show that $p_*$ is bijective more directly?
 A: *

*$p$ is bijective: In fact each null-homotopic map $f : S^1 \to S^1$ can be lifted to a map $\tilde f : S^1 \to \mathbb R$. Then the maps $\tilde f_k(z) = \tilde f(z) + k$ give us all lifts of $f$. Since $f(1) = 1$, we have $\tilde f(1) \in p^{-1}(1) = \mathbb Z$. Therefore $F  =\tilde f_{-\tilde f(1)}$ is the unique lift of $f$ such that $F(1) = 0$. This shows that $p_*$ is bijective.


*You are right, we have $p((-1/2,1/2))  = S^1 \setminus \{-1\}$. This was an error of the author. If you want, take $p(x) = \exp(\pi i x)$ to get $p((-1,1)) =  S^1 \setminus \{-1\}$ or leave it as it is and replace $(-1,1)$ by $(-1/2,1/2)$.


*Let us first see what we can say about $p_*(C(K,U))$ for subbasic open $C(K,U)$. In general we cannot immediately see that $p_*(C(K,U))$ is open. In fact, if $K \ne \emptyset$ and $U \ne \mathbb R$, then $C(K,U)$ is a proper subset of $Hom(S^1,\mathbb R)$ and thus $p_*(C(K,U))$ is a proper subset of $Hom(S^1,S^1)$. But if $U$ is too large, e.g. $U = (-1,1)$, then $C(K,p(U)) = C(K,S^1) =Hom(S^1,S^1)$, i.e. $p_*(C(K,U)) \subsetneqq C(K,p(U))$. A similar problem may arise for some $K$. As an example take $K = \{-1\}$ and $U = (-1/2,1/2)$. Then $p(U) = S^1 \setminus \{-1\}$. Define $f : S^1 \to S^1$ by $f(e^{2\pi it}) = e^{4 \pi it}$ for $t \in [0,1/2]$ and $f(e^{2\pi it}) = e^{4 \pi i(1-t)}$ for $t \in [1/2,1]$. This is an inessential map with $f(1) = f(-1) = 1$ (hence $f \in Hom(S^1,S^1)$ with  $f(K) \subset p(U)$). But the map $F = (p_*)^{-1}(f)$ which is the unique lift of $f$ such that $F(1) = 0$) is given by $F(e^{2\pi it}) = 2t$ for $t \in [0,1/2]$ and  $F(e^{2\pi it}) = 2(1-t)$ for $t \in [1/2,1]$. Therefore $F(-1) = F(e^{\pi i}) = 1 \notin U$, i.e. $F \notin C(K,U)$.
Of course the inclusion $p_*(C(K,U)) \subset C(K,p(U))$ holds always true. A sufficient condition for $p_*(C(K,U)) = C(K,p(U))$ is that $K$ is connected with $1 \in K$ and $U$ is contained in an open interval of the form $(a,a+1)$ with $a < 0 < a+1$. In fact, consider $f : S^1 \to S^1$ such that $f(K) \subset p(U)$. Let $V$ be the connected component of $0$ in $U$; this is an open interval of length $\le 1$. Since $p : U \to p(U)$ is a homeomorphism (recall $U \subset (a,a+1)$), we see that $p(V)$ is the connected component of $p(U)$ which contains $1$. We know that $f(K)$ is connected and contains $1$, thus $f(K) \subset p(V)$. The map $F = (p_*)^{-1}(f)$ is the unique lift of $f$ such that $F(1) = 0$. Thus $F(K) \subset p^{-1}(p(V))$. We conclude that that the connected set $F(K)$ is contained in the component of $p^{-1}(p(V))$ that contains $0 = F(1)$. This is $V$, thus $F  \in C(K,V) \subset C(K,U)$.
So what can be done to show that $p_*$ maps each open set to an open set?
The map $p_* : Hom(S^1,\mathbb R) \to Hom(S^1,S^1)$ is a group homomorphism, where the group structure is induced by addition $(F + G)(z) = F(z) + G(z)$ in $Hom(S^1,\mathbb R)$ and multiplication $(f \cdot g)(z) = g(z) \cdot g(z)$ in $Hom(S^1,S^1)$. Both $Hom$-groups are topological groups which means that the group operations and inversions are continuous. It is then well-known that it suffices to show that $p_*'$ is continuous resp. open at the neutral element which is the $0$-function. This means that it suffices to show that all "sufficiently small" open neighborhoods of $0$ are mapped to open sets. More precisely, we have to show that each open neighborhood $W$ of $0$ contains an open neighborhood $V$ of $0$ such that $p_*(V)$ is open. $W$ contains a basic open neigborhood of $0$ having the form $\bigcap_i C(K_i, U_i)$. Since we have $0 \in \bigcap_i C(K_i, U_i)$,  we get $0 = 0(K_i) \subset  U_i$. Hence $0 \in C(\bigcup_i K_i, \bigcap_i U_i) \subset \bigcap_i C(K_i, U_i)$. Let $K = \bigcup_i K_i$ and $U = (-1/2,1/2) \cap \bigcap_i U_i$. Then $0 \in V = C(S^1,U) \subset C(K,U) \subset C(\bigcup_i K_i, \bigcap_i U_i) \subset W$. This $V$ is that in your question.
Note that we cannot argue that it suffices to consider subbasic open $W$; recall the above considerations. We need to shrink $W$ to something which we know that it is mapped to an open set.
A: There was an answer by Paul Frost that helped me a good deal. I conclude that the proof in my textbook contains an error that I will try to fix in my third point below.

*

*The simplest argument that function is bijective indeed relies on the fact that every null homotopic function $S^1 \to S^1$ can be lifted and that all these lifts differ by an integer constant. It follows that there is a unique lift $\tilde{f}$ such that $\tilde{f}(0)=1$. Liftability of null homotopic functions can be derived from the Homotopy lifting property.

*We indeed may need to require $U\subseteq (-1/2, 1/2)$ for there to be $p(U) \subseteq S^1 \setminus \{-1\}$.

*Since $V,W$ are topological groups, it is enough to prove that every neighborhood of neutral element is mapped to a neighborhood of a neutral element; it can be then derived that all open sets are mapped to open sets. Further, since $p_*$ is bijective, and it holds that
$$p_* \left(\bigcap_i^{n} M(K_i, U_i)\right)=\bigcap_i^{n} p_* \left(M(K_i, U_i)\right),$$
we may want to show that every subbasic open neighbourhood $M(K, U)$ of $0$ contains subbasic neighbourhood $M(K^{\prime}, U^{\prime})$ of $0$ that is mapped to an open neighbourhood of neutral element $1$ in $H$. Since for any $K, U$ it's true that
$$M(K \cup S^1, (-1/2, 1/2) \cap U) \subseteq M(K, U),$$
it is sufficient to prove that $p_* M(S^1, U) = M(S^1, p(U))$ if we have $U \subseteq (-1/2, 1/2)$. We can prove it using by considering the following:
$$p_*^{-1} M(K, U) = M(K, p^{-1} U).$$
In our case we see that
$$p_*^{-1}M(S^{1}, p(U)) = M(S^{1}, p^{-1}p(U)) = M\left(S^{1}, \bigcup_{n \in \mathbb{Z}} n+U\right) \subseteq M\left(S^{1}, \bigcup_{n \in \mathbb{Z}} n+(-1/2, 1/2)\right).$$
The last inclusion is given so that we can emphasize that the open set is disjoint. Since we have that $S^1$ is connected, its image under $f \in Hom(S^1, \mathbb{R})$ is also connected. Since we also impose that $f(1)=0$, it is true that
$$M\left(S^{1}, \bigcup_{n \in \mathbb{Z}} n+U\right)=M\left(S^{1},U\right).$$
Finally, we've shown that
$$p_*^{-1}M(S^{1}, p(U))=M(S^{1},U),$$
from which follows $$p_*M(S^1, U) = M(S^1, p(U)).$$
It finishes my argument that an arbitrary neighborhood of $0$ contains a smaller open neighbourhood of $0$ that is mapped to an open neighbourhood of $1$.

