Context: A problem I'm solving asks "Prove that any two maps $S^m \rightarrow S^n$, where $n>m$ is homotopic. [Hint: use the Simplicial Approximation theorem]
My first thoughts were that if we have two maps $f,g$, since $S^n$ is path connected, I can construct a homotopy between any two maps by connecting the two points $f(x)$, $g(x)$ via some path. But the hint makes me think I'm missing something.
To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y \to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = \operatorname{id}_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m \to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m \to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
Not necessarily.
Take $f: S^n \to S^n$ as identity map And. $g: S^n \to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected. But it is contradicting the fact that $S^n$ is not contractible
