Let $f(x)$ and $g(x)$ be two continuous functions on $[0,1]$ and $$\int_{0}^{1}f(x) dx= \int_{0}^{1}g(x)dx = 1$$

Show that there exist $[a,b]\subset [0,1]$, such that $$\int_{a}^{b}f(x) dx= \int_{a}^{b}g(x)dx = \frac{1}{2} $$

The question can be solved by considering the fundamental group of $S^1$, now I am wondering if we can solve it by real-analysis.

Here is the topological solution:

Assume that for any $[a,b]\subset[0,1]$, we always have $$(\int_{a}^{b}f(x) dx\neq \frac{1}{2}) \quad \vee \quad(\int_{a}^{b}g(x)dx \neq \frac{1}{2}) $$ Consider mapping $$\phi:D\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\},\,\,(x,y)\mapsto(\int_{y}^{x}f(t) dt,\int_{y}^{x}g(t) dt)$$ where $D=\{(x,y)|0\leq x\leq y\leq 1\}$.

Let $a$ be path from $(0,0)$ to $(0,1) $in $D$ and $b$ be path from $(0,1)$ to $(1,1) $in $D$, then $ab$ is a path from $(0,0)$ to $(1,1) $in $D$ and $$\phi\circ (ab):[0,1]\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\} $$ Notice that $$(\phi\circ (ab))(0)=(\phi\circ (ab))(1)=(0,0)$$

so $\phi\circ (ab)$ is a loop based on $(0,0)$ in $\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}$.

For any $t\in [0,\frac{1}{2}]$, it's not hard to get that $$(\phi\circ (ab))(t+\frac{1}{2})=(1,1)-(\phi\circ (ab))(t)$$ is equivalent to $$(\phi\circ (ab))(t+\frac{1}{2})-(\frac{1}{2},\frac{1}{2})=-((\phi\circ (ab))(t)-(\frac{1}{2},\frac{1}{2}))$$

Define retraction $$r:\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}\rightarrow S^1,\quad (x,y)\rightarrow\ \frac{(x,y)-(\frac{1}{2},\frac{1}{2})}{||(x,y)-(\frac{1}{2},\frac{1}{2})||}$$

Then $r\circ \phi\circ (ab)$ is a loop on $S^1$, such that $$(r\circ \phi\circ (ab))(t+\frac{1}{2})=-(r\circ \phi\circ (ab))(t),\quad\forall t\in [0,\frac{1}{2}]$$

So $<r\circ \phi\circ (ab)>$ is not trivial in $\pi_1(S^1)$.

However, $ab$ is path homotopic to $c$ in $D$, where $c$ is the path between $(0,0)$ and $(1,1)$ in $D$. In this case, $r\circ \phi\circ (ab)$ is a point-path in $S^1$ by $$(\phi\circ c)(t)=\phi(t,t)\equiv (0,0)$$

which leads to contradiction.

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    Just for curiosity, what's the argument using the fundamental group? – Dante Grevino Dec 12 '18 at 07:28
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    @DanteGrevino Hi, I add the argument. –  Dec 14 '18 at 03:48
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    Hi. It is a nice argument, thanks! I will think of a solution using real-analysis in my free-time – Dante Grevino Dec 14 '18 at 04:44
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    @Lau: I wonder if your problem is related (or even equivalent) to https://math.stackexchange.com/q/804150/42969, where a similar argument as your's is used in one of the answers. – Martin R Dec 15 '18 at 18:28
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    @MartinR I think the two arguments use the same technique, but I can't clearly see the relationship between them. –  Dec 16 '18 at 02:48
  • I remember reading a paper in a popular math magazine with this problem solved by means of the Brouwer Fixed Point Theorem. It was in the late 1990's. – Peter Saveliev Dec 18 '18 at 19:02
  • @PeterSaveliev We can also use fundamental group to prove BFPT, so the argument in that paper may be related to the one I give. –  Dec 19 '18 at 00:16
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    @Lau I've done some searching, it's not BFPT, but rather this: a map from disk to disk, boundary to boundary, with deg≠0 is onto (Kronecker?). The paper is by Vilmos Totik in American Mathematical Monthly, March 1999. – Peter Saveliev Dec 19 '18 at 02:27
  • @PeterSaveliev Thanks a lot, I will look into it. –  Dec 19 '18 at 02:29
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    The paper PeterSaveliev mentioned: https://www.jstor.org/stable/pdf/2589678.pdf –  Dec 21 '18 at 02:04
  • It should be mentioned for clarity that the paths $a,b,c$ in $D$ are all straight line segments. – Qmechanic Dec 24 '18 at 15:27

4 Answers4


Consider the following proposition.

Proposition 1: Let $T:=\{(a,b)\in[0,1]^2:a\leq b\}$ and let $f:T\to\mathbb{R}^2$ be continuous. Assume that $f(0,r)+f(r,1)=(1,1)$ holds for all $r\in[0,1]$ and that $f$ is $(0,0)$ on the edge $E:=\{(a,b)\in T:a=b\}$. Then there exists $v\in T$ such that $f(v)=(\frac12,\frac12)$.

In can easily be seen that your statement follows from Proposition 1 by considering $$F(a,b)=\left(\int_a^bf(x)\mbox{d}x,\int_a^bg(x)\mbox{d}x\right).$$ I think it is unlikely that you can prove your statement with a proof that can not be manipulated to also prove Proposition 1. However, Proposition 1 can be shown equivalent to a version of the Borsuk Ulam theorem, only using arguments from real analysis. Since I don't think anyone ever came up with a proof of the Borsuk Ulam theorem only using arguments from real analysis, I think it is unlikely that you can prove your statement only using arguments from real analysis. The version of the Borsuk Ulam theorem I am talking about is the following.

Borsuk Ulam theorem: Let $f:D^2\to\mathbb{R}^2$ be continuous such that $f(v)=-f(-v)$ for $v\in S^1$. Then there exists $v\in D^2$ such that $f(v)=(0,0)$.

Remarks: Here $D^2=\{v\in\mathbb{R}^2:\|v\|\leq1\}$ and $S^1=\{v\in\mathbb{R}^2:\|v\|=1\}$. A generalized form of this theorem is mentioned equivalent to the Borsuk Ulam theorem in the second bullet point of the first paragraph of the Wikipedia page. However, from now on, I will refer to this as the Borsuk Ulam theorem.

I think it is worth mentioning that, even though the Borsuk Ulam theorem is usually proven with algebraic topology, you can also prove the Borsuk Ulam theorem with just combinatorics and real analysis. At the end of this post, I give a direct proof of the Borsuk Ulam theorem, only using Tucker's lemma from combinatorics and the Bolzano Weierstrass theorem from real analysis.

The rest of this post will be showing the equivalence of Proposition 1 and the Borsuk Ulam theorem. The following proposition will be used in the proof of both implications.

Proposition 2: There exists a homeomorphism $\phi:T\setminus E\to D^2\setminus\{(1,0)\}$ such that $\phi(0,r)=e^{ri\pi}$ for all $r\in(0,1]$ and $\phi(r,1)=-e^{ri\pi}$ for all $r\in[0,1)$.

Remark: For the definitions of $T$, $E$ and $D^2$, refer to Proposition 1 and the Borsuk Ulam theorem.

If you want, you can ask in the comments for an explicit proof of this proposition. However, the proof will be very tedious and not very enlightening at all. The reader is advised to draw pictures of the situation and convince themselves that such a homeomorphism really exists.

Borsuk Ulam $\implies$ Proposition 1

Let $f$ be given as in Proposition 1. Let $\phi$ be as in Proposition 2. We can define $g:D^2\to\mathbb{R}^2$ by \begin{equation} g(v)= \begin{cases} (0,0) & \mbox{if $v=(1,0)$} \\ f(\phi^{-1}(v)) & \mbox{otherwise} \end{cases}. \end{equation} Combining the properties of $f$ and $\phi$, we know that $g$ is continuous, and $g(v)+g(-v)=(1,1)$ holds for all $v\in S^1$. We find that $g-(\frac12,\frac12)$ obeys all conditions of the Borsuk Ulam theorem. Hence, there exists $v\in D^2$ such that $g(v)=f(\phi^{-1}(v))=(\frac12,\frac12)$.

Proposition 1 $\implies$ Borsuk Ulam

Let $f:D^2\to\mathbb{R}^2$ be as in the Borsuk Ulam theorem. The goal is to find $v\in D^2$ such that $f(v)=(0,0)$. It is a simple application of the intermediate value theorem to show there exists $v\in S^1$ such that $f(v)$ has the same $x$ and $y$ coordinate. Note that this means either $f(v)=(0,0)$, or one of $f(v)$ and $f(-v)$ has equal negative coordinates. In the first case, we are done. In the second case, we can assume without loss of generality (by scaling and rotating) that $f(1,0)=(-\frac12,-\frac12)$.

We can define $g:T\to\mathbb{R}^2$ by \begin{equation} g(v)= \begin{cases} (0,0) & \mbox{if $v\in E$} \\ f(\phi(v))+(\frac12,\frac12) & \mbox{otherwise} \end{cases}. \end{equation} Combining the properties of $f$ and $\phi$, we know that $g$ obeys all conditions of Proposition 1. Hence, there exists $v\in T$ such that $g(v)=(\frac12,\frac12)$, so $f(\phi(v))=(0,0)$.

Combinatorial proof of the Borsuk Ulam theorem

Since $D^2$ is compact, $f$ is uniformly continuous. So for all $\varepsilon>0$ there exists $\delta>0$ such that for all $v,w\in D^2$ with $\|v-w\|<\delta$ we have $\|f(v)-f(w)\|<\varepsilon$.

Let $T$ be a triangulation of $D^2$ such that all edges have length smaller than $\delta$. We then give every vertex a color from the set $\{1,-1,2,-2\}$. The color of vertex $v$ is purely based on $f(v)$. If $f(v)$ has an absolutely larger first coordinate than second coordinate, $v$ gets color $1$ or $-1$, depending on the sign of the first coordinate of $f(v)$. Otherwise, $v$ gets color $2$ or $-2$, depending on the sign of the second coordinate of $f(v)$.

By the given properties of $f$ we can apply Tucker's lemma. We find adjacent vertices $v,w\in D^2$ with opposite colors. Because of our choice in triangulation, we find $\|f(v)-f(w)\|<\varepsilon$. Then because of our way of coloring the vertices, we see that both coordinates of both $f(v)$ and $f(w)$ must be absolutely smaller than $\varepsilon$, so in particular $\|f(v)\|<2\varepsilon$.

We find for each $\varepsilon>0$ some $v\in D^2$ such that $\|f(v)\|<2\varepsilon$. Hence, we can find a sequence $\{v_i\}$ in $D^2$ such that $f(v_i)\to(0,0)$. Because $D^2$ is compact, by the Bolzano Weierstrass theorem there is a convergent subsequence of $\{v_i\}$ with limit $v\in D^2$. By the continuity of $f$, we find $f(v)=(0,0)$.

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Here is a proof for the case $f > 0$.

Let $A$ satisfy $$ \int_0^A f(x) \, dx = \frac{1}{2}. $$ For every $a \leq A$, there exists a minimal point $\beta(a) > a$ such that $$ \int_a^{\beta(a)} f(x) \, dx = \frac{1}{2}. $$ Note that $\beta(0) = A$ and $\beta(A) = 1$.

Define $$ G(a) = \int_a^{\beta(a)} g(x) \, dx, $$ and notice that $$ G(0) + G(A) = \int_0^A g(x) \, dx + \int_A^1 g(x) \, dx = \int_0^1 g(x) \, dx = 1. $$ Therefore either $G(0) \geq 1/2 \geq G(A)$ or $G(0) \leq 1/2 \leq G(A)$. Either way, since $G$ is continuous, there must exist a point $a \leq A$ such that $G(a) = 1/2$. Taking $b = \beta(a)$, we obtain $$ \int_a^b f(x) \, dx = \int_a^b g(x) \, dx = \frac{1}{2}. $$

Yuval Filmus
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    I probably missed something but why is $G$ continuous? – BigbearZzz Dec 14 '18 at 09:04
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    In order for $G$ to be continuous you need that $a \to \beta(a)$ is continuous. That is not obvious to me, in particular since $\int_a^{\beta(a)} f(x) \, dx = \frac{1}{2}$ does not define $\beta(a)$ uniquely. – Martin R Dec 14 '18 at 09:05
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    These are all technicalities. This sketch can probably be converted to a bona fide proof with some effort. – Yuval Filmus Dec 14 '18 at 09:08
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    @MartinR $\beta$ is not continuous for $f=2\chi_{[0,1/3]}- \chi_{[1/3,2/3]}+2\chi_{[2/3,1]}$ but this $f$ is not continuous. I think by smoothing this $f$ we can get a counter example. – BigbearZzz Dec 14 '18 at 09:09
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    @YuvalFilmus I think the problem Martin addressed is actually genuine, not some technicality. – BigbearZzz Dec 14 '18 at 09:09
  • The problem only happens since $f,g$ could vanish. This suggests defining $\beta_-(a),\beta_+(a)$ as the minimum and maximum values that work, and seeing if this fix can be made to work. – Yuval Filmus Dec 14 '18 at 09:12
  • Your proof seems fine if we also assume that $f,g\ge 0$. The (mollification of) $f$ I gave earlier is a counter example showing that $\beta_{-}$ cannot be continuous. – BigbearZzz Dec 14 '18 at 09:15
  • Do you like my updated argument? If it works, then this vindicates my claim that the problem was just a technicality. Though admittedly, I was secretly considering the case of $f,g \geq 0$ (i.e. the ham sandwich theorem). – Yuval Filmus Dec 14 '18 at 09:19
  • This still doesn't work if $f$ can take negative values imo. Can you prove that it works? I don't think it would but I could be mistaken. – BigbearZzz Dec 14 '18 at 09:25
  • Ok, I think we all agree it should work when, say, $f > 0$. Hopefully someone will be able to patch it for the general case. – Yuval Filmus Dec 14 '18 at 09:27
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    @BigbearZzz For an example of a continuous function $f$ with $\beta$ not continuous, we can have $f$ start at 0, be a positive triangle with area $\frac{1}{2}$, then be a negative triangle with area, say, $\frac{1}{4}$, and then be a positive triangle with area $\frac{3}{4}$. The point is that $\beta(0)$ will be the right end of the first triangle, while, for each $n \ge 1$, $\beta(\frac{1}{n})$ will be somewhere on the third triangle. (by "triangle", I just mean $f$ is piecewise linear with alternating signed sloped) – mathworker21 Dec 19 '18 at 23:19
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    I give my award to the answer for it really made the question noticed. –  Dec 21 '18 at 09:50
  • If $f>0$, $\beta$ is locally defined and continuous (and even $C^1$) by the implicit function theorem. Since $\beta$ is uniquely defined in the case $f>0$, it is globally defined and continuous everywhere. – Aloizio Macedo Dec 21 '18 at 15:41

Since my math is a bit rusty, I do my best. Feel free to correct and/or criticise.

Imagine the following two functions:

$$F(a,b)=\int_a^b f(x)\,\textrm{d}x$$


$$G(a,b)=\int_a^b g(x)\,\textrm{d}x$$

of which we know that $F(0,1)=G(0,1)=1$ and $F(x,x)=G(x,x)=0$ and both $F(x,y)$ and $G(x,y)$ are continious since $f(x)$ and $g(x)$ are continious.

We know the following:

  1. There exists a set $\mathcal{A}=\{\alpha_1 \le \alpha_2 \le \ldots \le \alpha_n\}$ such that $F(0,\alpha_i)=1/2$. If $\alpha$ is the solution to $F(0,\alpha)=1/2$ and has multiplicity $k$, than it will appear $k$ times in the set $\mathcal{A}$.
  2. If $F(0,\alpha_i)=1/2$ than $F(\alpha_i,1)=1/2$ as $F(0,1)=F(0,\alpha_i)+F(\alpha_i,1)=1$
  3. $F(\alpha_1,\alpha_2)=F(\alpha_2,\alpha_3)=\ldots=F(\alpha_{n-1},\alpha_n)=0$. This comes from the fact that $F(0,\alpha_i) = F(0,\alpha_1) + F(\alpha_1,\alpha_i) = 1/2$
  4. Due to continuity of $F(x,y)$, we know that there must exist at least one contour $\mathcal{C}_f:=F(x,y)=1/2$ which connects the points $F(0,\alpha_i)$ with $F(\alpha_j,1)$.

  5. We can make the following table indicating if $F(0,x)\substack{>\\<}1/2$ and $F(x,1)\substack{>\\<}1/2$

       x     0     a1    a2    a3   a(n-1) an    1
    F(0,x)   0  <  =  >  =  <  = ... =  <  =  >  1
    F(x,1)   0  >  =  <  =  >  = ... =  >  =  <  1

    This table takes the multiplicity of $\alpha$ into account. It, thus also holds that $n$ must be odd!

The same holds for $G(x,y)$ with a set $\{\beta_1 < \ldots < \beta_m\}$.

The following figures show a couple of such contours for a set of functions:

enter image description here

The left image shows 4 functions satisfying the conditions of (a) continuous and (b) $\int_0^1 f(x)\textrm{d} x=1$. The right image shows the corresponding contours $F(x,y)=1/2$. For blue and orange, there is only a single contour and a single $\alpha$, for the green and red curve, there are several $\alpha$. To understand this image you know that all points on the Y-axis represent the integrals $\int_0^b f(x)\textrm{d}x$, while all points on top represent $\int_a^1 f(x)\textrm{d}x$. Any point with coordinates $(a,b)$ on the contour implies $\int_a^b f(x)\textrm{d}x=1/2$. The regions where $F(a,b)=\int_a^b f(x)\textrm{d}x>1/2$ are hatched, the gray area is invalid as we assume $a<b$.

enter image description here Here, the left image shows the curves $F(0,x)$ and the right $F(x,1)$. It is clear that everytime $F(0,x)$ is greater than $1/2$, $F(x,1)$ is smaller than $1/2$ and vice versa.

Assume $n=m=1$ with $\alpha_1<\beta_1$: It is clear that both contour lines $\mathcal{C}_f$ and $\mathcal{C}_g$ must intersect. Since $\alpha_1<\beta_1$, the point $(0,\beta_1)$ lays in the region where $F(x,y)>1/2$ while the point $(\beta_1,1)$ lays in the area where $F(x,y)<1/2$. Hence due to continuity, both curves must intersect, demonstrating that there exists a region $[a,b]$ such that

$$\int_a^b f(x)\,\textrm{d}x = \int_a^b g(x)\,\textrm{d}x = \frac{1}{2}$$

Assume $n>1$ and $m=1$: Again with the same reasoning we can conclude that there must exist a point $(a,b)$ such that $F(a,b)=G(a,b)=1/2$. Imagine $\alpha_i<\beta_1<\alpha_{i+1}$ and assume $F(0,\beta_1)<1/2$, than we know from point 5 that $F(\beta_1,1)>1/2$. Now, since $m=1$ and $G(x,y)$ is continious we know that there is a path $\mathcal{C}_g:=G(x,y)=1/2$ connecting $(0,\beta_1)$ with $(\beta_1,1)$. Projecting this path onto $F(x,y)$ thus implies that somewhere there must exist a $(a,b)$ where $F(a,b)=G(a,b)=1/2$ or

$$\int_a^b f(x)\,\textrm{d}x = \int_a^b g(x)\,\textrm{d}x = \frac{1}{2}$$

Assume $n>1$ and $m>1$: Still, need to think about this one ... I think it can be proved similarly to the previous two cases which use point (5).

A few cases I can imagine, are presented in the next figure:

  • Contours $\mathcal{C}$ which are running through regions always above 1/2 or below 1/2 (dashed blue line vs orange region)
  • Contours which are always contained in such regions (dashed orange line vs orange region)

but there must be many other configurations. The second point can be countered directly using point (4) of above. If $F(0,\beta)>1/2$ than $F(\beta,1)$ must be less than $1/2$. So the second cannot be true.

enter image description here

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    Thanks to the good job you did in $m=n=1$ and those graphs. ( PS: the intersection of the cruve is not trivial due to https://math.stackexchange.com/questions/804150/prove-that-two-paths-on-opposing-corners-of-the-unit-square-must-cross ) –  Dec 21 '18 at 02:18
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    The pictures look pretty impressive to be honest. – BigbearZzz Dec 21 '18 at 15:17

OP's problem can be slightly generalized as follows:

Let there be given two continuous functions $F_1,F_2:[0,1]\to \mathbb{R}$ with $$F_1(0)~=~0~=~F_2(0)\quad\text{and}\quad F_1(1)~=~1~=~F_2(1).\tag{1}$$ Define the 2-simplex/triangle $$\Delta~:=~\{(x,y)\in\mathbb{R}^2| 0\leq x \leq y\leq 1\}.\tag{2}$$ Define a function ${\cal F}=({\cal F}_1,{\cal F}_2): \Delta\to \mathbb{R}^2$ by its components $${\cal F}_1(x,y)~:=~ F_1(y)-F_1(x)\quad\text{and}\quad{\cal F}_2(x,y)~:=~ F_2(y)-F_2(x).\tag{3}$$ Then the statement is $$\exists (a,b)\in \Delta: ~~{\cal F}(a,b)~=~(\frac{1}{2},\frac{1}{2}).\tag{4} $$

It is natural to ponder if the point $(\frac{1}{2},\frac{1}{2})$ in the statement (4) can be replaced with another point [apart from the trivial points $(0,0)$ and $(1,1)$]? Simple counterexamples say no. We shall see in eq. (8) below that the point $(\frac{1}{2},\frac{1}{2})$ indeed is a special point.

Let $\alpha, \beta, \gamma:[0,1]\to \Delta$ be three curves along the boundaries of $\Delta$: $$\alpha(t)~:=~(0,t) ,\qquad \beta(t)~:=~(t,1) ,\qquad \gamma(t)~:=~(t,t), \qquad t~\in~[0,1]. \tag{5}$$

We calculate that $${\cal F}\circ \alpha(t)~=~(F_1(t),F_2(t)), \qquad {\cal F}\circ \beta(t)~=~(1-F_1(t),1-F_2(t)), \qquad {\cal F}\circ \gamma(t)~=~(0,0), \tag{6}$$ and $${\cal F}\circ \alpha(0)~=~(0,0)~=~{\cal F}\circ \beta(1), \qquad {\cal F}\circ \alpha(1)~=~(1,1)~=~{\cal F}\circ \beta(0). \tag{7}$$ In particular, the average curve $$ \frac{1}{2}\left\{ {\cal F}\circ \alpha(t) + {\cal F}\circ \beta(t)\right\}~=~(\frac{1}{2},\frac{1}{2}) \tag{8} $$ is a constant curve! To perform an indirect proof we next assume that the statement (4) is wrong. Eqs. (7) & (8) then imply that the ${\cal F}$-image of the concatenation $\alpha\#\beta$ is a loop around $(\frac{1}{2},\frac{1}{2})$ with odd winding number. This clashes with the fact that the two curves $\alpha\#\beta$ and $\gamma$ are homotopic, cf. OP's topologic proof. See user SmileyCraft's answer for proofs via Borsuk-Ulam theorem and Tucker's lemma.

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  • Even though it is intuitively clear that the winding number is odd, do you have a proof of this? Preferably only using arguments from real analysis obviously. Although I believe the winding number isn't even a concept from real analysis. – SmileyCraft Dec 31 '18 at 15:30