# Can all induced maps be described categorically.?. (or at least as generally as possible)

Hi: I am new here. I went over the fAQ's, still, sorry if I break protocol.

I am pretty confused about induced maps in different areas of algebraic topology; I do know how these induced maps are defined in many cases, but I definitely do not understand well-enough the rules governing when a map between two topological spaces X,Y , induces a map in homology, or homotopy. AFAIK, if we have a map f:X-->Y , and this map takes cycles to cycles and boundaries to boundaries, then this map "passes to homology" (not clear what that means.).
Problem(at least to me) is that this word "induced" seems to be overused (in the sense that its meaning does not always seem clear.): induced quotients, induced homomorphisms, induced bundles, etc. So: does anyone know if induced maps can be described categorically, or at least, could someone please explain when a given map between topological spaces induces a map on homology or cohomology.?.

I think there is some underlying algebraic result dealing with normal subgroups(which extends to any subgroup in homology, since chain groups are Abelian.), but I am not too sure of this.

Thanks For any Help.

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The key word in this context is functor. The point is that homology, homotopy etc. are functors. For example consider homology $H_n$. This is a functor from the category of topological spaces to the category of Abelian groups. In categories, although the usual notation obscures it, morphisms are more important than objects. It is crucial to define the action of the functor on morphisms. Returning to our example, to define the functor $H_n$ we need to define an Abelian group $H_n(X)$ for each topological space $X$, and for each continuous map $f:X\to Y$ a group homomorphism $H_n(f):H_n(X)\to H_n(Y)$. These maps $H_n(f)$ must preserve composition and identities. In practice, we don't use the notation $H_n(f)$ for these maps but typically use alternatives like $f_*$ which is quicker to write, but less informative. Similarly homotopy groups and cohomology groups form functors on suitable categories.

There is a whole algebra of categories, functors and more which are dealt with in texts on category theory. For instance the composition of two functors is a functor. In our example, some texts see the homology group functor as a composite, being a functor from topological spaces to chain complexes, and the functor taking a chain complex to its homology groups. Also texts on topology vary in the detail in which they explain the construction of the maps I've denoted as $H_n(f)$; some go into lots of detail while other wave their hands more. In general once they have done some examples in detail, they tend not to go into so much detail in subsequent examples, as they assume that the reader can now fill in more details.

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Thanks : I have a chicken-egg confusion here when going beyond (co)homology, tho:I know that we define (Eilenberg-Steenrod) (Co)homology to be a functor;there are other cases,tho, in which we may not know in advance whether we have a functor:given any linear map between v.spaces V,W, we get a map W*-->V* (I think V->V* is a functor) . Given a map between manifolds M,N , we get an induced map between the respective tangent spaces T_pM, T_pN (where is the functor.?). Do these induced maps (all other) also follow from functoriality.?. Also, once we have functoriality, how to define f*? thanks. – confused May 23 '10 at 4:55
You are raising a lot of questions here. The Eilenberg-Steenrod axioms do not define of a (co)homology functors, they axiomatize them. Dualisation of vector spaces is a contravariant functor. You did not mention manifolds originally, but a map between two smooth manifolds induces vector maps from $T_p(M)$ to $T_{f(p)}(M)$ which can be regarded as a functor. The domain category is the category of manifolds with a base point, with base-point preserving maps, and the codomain category is that of vector spaces – Robin Chapman May 23 '10 at 10:32

Let me add that the answer to your question "when does a map between spaces induce a map in homology/cohomology/homotopy?" is "always" (as long as you stick to continuous maps..).

In fact (say for homology) if you have a continuous $f:X\to Y$, the induced homomorphism of chain complexes $C_\bullet(f):C_\bullet(X)\to C_\bullet(Y)$ sends automatically cycles to cycles and boundaries to boundaries (simply because it is compatible with the differentials of the two complexes, in the sense that $f\circ d_X=d_Y\circ f$).

The fact that it "passes" to homology is now an algebraic fact, namely the fact that if you have four abelian groups $A,B,C,D$ and three homomorphisms $f:A\to B, g:A\to C, h:B\to D$ with $g$ and $h$ surjective (so $C$ is a quotient of $A$ and $D$ is a quotient of $B$), then you can find $f':C\to D$ such that $f'\circ g=h\circ f$ if and only if $f(\ker(g))\subseteq \ker(h)$. (you might want to draw a diagram here :D)

Take $A=Z_n(X), B=Z_n(Y), C=H_n(X)=Z_n(X)/B_n(X), D=H_n(Y)=Z_n(Y)/B_n(Y)$, where $Z_\bullet=$cycles and $B_\bullet=$ boundaries, as usual, and you get your induced map in homology.

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Sorry, but my algebra is a bit weak(i am a begginer, please be patient): i know that a map passes to the quotient if it is constant in equivalent classes;I know the condition needed to make triangular diagrams commute, but I do not understand the precise meaning of "passing to homology": you mention that cycles are sent to cycles, but I don't see a amp sending Bn(X) to Bn'(Y). Would you please define the meaning of "passing to homology" or give a ref.?. I know it must have to see with:1)f(B_n(X))<B_n(Y) and 2) f(Z_n(X))<(Z_n(Y) and 3)your condition on kernels. Then what.? Thanks. – confused May 23 '10 at 4:59
"passing to homology" is just a way to say that the various maps C_n(f):C_n(X)->C_n(Y) induce maps in homology H_n(f):H_n(X)->H_n(Y), as in my answer above. it is defined simply like this: given an element a of H_n(X) you take a cycle c in Z_n(X) such that a=[c] in H_n(X)=Z_n(X)/B_n(X), then you apply C_n(f) to c and take its class in H_n(Y)=Z_n(Y)/B_n(Y). the facts that you mentioned above assure that this is well-defined. – Mattia Talpo May 23 '10 at 20:47