If $s:|K^m|\to |L|$ is a simplicial approximation to $f:|K^m|\to |L|$ and $t:|L^n|\to |M|$ is a simplicial approximation to $g:|L^n|\to |M|$. Is $t\circ s: |K^{m+n}|\to |M|$ necessarily a simplicial approximation to $g\circ f:|K^{m+n}|\to |M|?$
I'm sure answer is yes. But i don't know how to prove this. This is one of the question in final exam but i didn't do. Most probably it again comes in my qualifying exam so please I seek the help. Your help will be appreciated. Thanks in Advance.
If I did not overlook this too quickly, the answer should be yes. We know that, by definition, $s: |K| \to |L|$ is a simplicial approximation to $f: |K| \to |L|$ if $s(x)$ lies in the carrier of $f(x)$ for all $x \in |K|$.
Now, given $f,g,s,t$ as in your question, pick some $x \in |K^{m+n}| = |K|$. Let $f(x) = y$ and $s(x) = y'$. Note that $y$ by definition is not necessarily equal to $y$. What we know is that they do lie in the same interior of some simplex $A$ of $L^n$. Now, we ask if $t\circ s(x)$ is a simplicial approximation to $g\circ f(x)$; that is, we ask that $g(y)$ and $t(y')$ lie in the same interior of some simplex of $M$. We now that $t$ is a simplicial map, i.e. if $A$ is a simplex of $L^n$, $t(A)$ is a simplex of $M$. Therefore, since $y$ and $y'$ both lie in $A$, $t(y),t(y') \in t(A)$. But since $t$ approximates $g$, this means that $g(y) \in t(A)$. This shows that the composition $t\circ s$ is a simplicial approximation to $g\circ f$