Every simplicial map is cellular.

A simplicial map is by definition a map $$f:K\to L$$ between simplicial complexes that sends each simplex of $$K$$ to a simplex of $$L$$ by a linear map taking vertices to vertices.

A cellular map is by definition a map $$f:X\to Y$$ between cellular complexes such that $$f(X^n)\subset Y^n$$ for all $$n$$, where $$X^n$$ is the $$n$$-th skeleton of $$X$$.

Is it true that every simplicial map is cellular? I think it seems true but I'm not sure.

• Your definition does not seem standard to me. Take a simplicial complex and then make it finer. Is the inclusion from the finer complex to the first complex simplicial? – s.harp Mar 12 at 21:20
• @s.harp: presumably by "making a simplicial complex finer" you mean "subdividing simplices". If so, then there is no "inclusion" from the finer complex to the coarser complex. (However, there is a simplicial map.) – Rob Arthan Mar 12 at 22:21
• @s.harp This definition is from Hatcher's. – probably123 Mar 12 at 23:20

Hint: by definition, a simplicial map $$f : K \to L$$, maps the vertices of an $$i$$-simplex $$\sigma$$ of $$K$$ onto a set of vertices in $$L$$ that span a $$j$$-simplex $$\tau$$ of $$L$$, where necessarily $$j \le i$$. So $$f$$ maps the $$i$$-skeleton of $$K$$ (which comprises the union of the $$k$$-simplices in $$K$$ with $$k \le i$$) into a union of $$l$$-simplices in $$L$$ with $$l \le i$$, which is contained in the $$i$$-skeleton of $$L$$ (which comprises the union of all the $$l$$-simplices in $$L$$ with $$l \le i$$).