every map into sphere which is not onto is nullhomotopic

Proposition. Let $X$ be a topological space and $f:X\to {𝕊}^n$ a continous map from $X$ to $n$-dimensional sphere which is not onto. Then $f$ is nullhomotopic.

Proof. Assume that there is $y_0\in {𝕊}^n$ such that $y_0\notin \mathrm{im}(f)$. It is well known that there is a homeomorphism $\varphi :{𝕊}^n\setminus \{y_0\}\to {ℝ}^n$. Then we have an induced map

$$\varphi \circ f:X\to {ℝ}^n.$$

Since ${ℝ}^n$ is contractible, then there is $c\in {ℝ}^n$ such that $\varphi \circ f$ is homotopic to the constant map in $c$ (denoted with the same symbol $c$). Let $\psi :{ℝ}^n\to {𝕊}^n$ be a map such that $\psi (x)={\varphi }^{-1}(x)$ (note that $\psi$ is not the inverse of $\varphi$ because $\psi$ is not onto) and take any homotopy $H:\mathrm{I}\times X\to {ℝ}^n$ from $\varphi \circ f$ to $c$. Then we have a homotopy $F:\mathrm{I}\times X\to {𝕊}^n$ defined by the formula $F=\psi \circ H$. It is clear that

$$F(0,x)=\psi (H(0,x))=\psi (\varphi (f(x)))=f(x);$$

$$F(1,x)=\psi (H(1,x))=\psi (c)\in {𝕊}^n.$$

Thus $F$ is a homotopy from $f$ to a constant map. $\mathrm{\square }$

Corollary. If $A\subseteq {𝕊}^n$ is a deformation retract of ${𝕊}^n$, then $A={𝕊}^n$.

Proof. If $A\subseteq X$ then by deformation retraction (associated to $A$) we understand a map $R:\mathrm{I}\times X\to X$ such that $R(0,x)=x$ for all $x\in X$, $R(1,a)=a$ for all $a\in A$ and $R(1,x)\in A$ for all $x\in X$. Thus a deformation retract is a subset $A\subseteq X$ such that there is a deformation retraction $R:\mathrm{I}\times X\to X$ associated to $A$.

Assume that $A$ is a deformation retract of ${𝕊}^n$ and $A\ne {𝕊}^n$. Let $R:\mathrm{I}\times {𝕊}^n\to {𝕊}^n$ be a deformation retraction. Then $r:{𝕊}^n\to {𝕊}^n$ such that $r(x)=R(1,x)$ is homotopic to the identity map (by definition of a deformation retract), but on the other hand it is homotopic to a constant map (it follows from the proposition, since $r$ is not onto, because $A$ is a proper subset of ${𝕊}^n$). Thus the identity map is homotopic to a constant map, so ${𝕊}^n$ is contractible. Contradiction. $\mathrm{\square }$