 way we can express the fact that "boundaries of (n+1)-cycles correspond to the absence of holes", besides per fiat?

The graph $X_1$: 
Let us now enlarge the preceding graph $X_1$ by attaching a 2-cell $A$ along the cycle $a-b$, producing a 2-dimensional cell-complex $X_2$. If we think of the 2-cell $A$ as being oriented clockwise, then we can regard its boundary as the cycle $a-b$. This cycle is now homotopically trivial since we can contract it to a point by sliding over $A$. In other words, it no longer encloses a hole in $X_2$. This suggests that we form a quotient of the group of cycles in the preceding examples by factoring out the subgroup generated by $a-b$. In this quotient the cycles $a-c$ and $b-c$, for example, become equivalent, consistent with the fact that they are homotopic in $X_2$.
$X_2$: 
Algebraically we can define now a pair of homomorphisms $C_2 \overset{\partial_2}{\to} C_1 \overset{\partial_1}{\rightarrow} C_0$ where $C_2$ is the infinite cyclic group generated by $A$ and $\partial_2(A)=a-b$. The map $\partial_1$ is the boundary homomorphism in the previous example. The quotient group we are interested in is Ker$\partial_1$/Im $\partial_2$, the kernel of $\partial_1$ module the image of $\partial_2$, or in other words, the 1-dimensional cycles modulo those that are boundaries [of 2-dimensional cycles], the multiples of $a-b$. The quotient group is the homology group $H_1(X_2)$... In the present example $H_1(X_2)$ is free abelian on two generators, $b-c$ and $c-d$, expressing the geometric fact that by filling in the 2-cell $A$ we have reduced the number of 'holes' in our space from three to two.

My question being: how do we express the motivation for modding/quotienting/factoring out the boundaries in topological terms? In geometric terms?

A:

This is just one example. In this example, there is only one 2
way we can express the fact that " boundaries of (n+1)-cycles correspond to the absence of hole ", besides per decree? 

 The graph $ X_1 $: 
 Let us now blow up the precede graph $ X_1 $ by attaching a 2 - cell $ A$ along the bicycle $ a - b$, produce a 2 - dimensional cell - complex $ X_2$. If we remember of the 2 - cell $ A$ as being oriented clockwise, then we can involve its boundary as the cycle $ a - b$. This bicycle is nowadays homotopically trivial since we can contract it to a point by sliding over $ A$. In other parole, it no longer encloses a hole in $ X_2$. This suggests that we imprint a quotient of the group of cycles in the preceding examples by factor out the subgroup generated by $ a - b$. In this quotient the cycles $ a - c$ and $ b - c$, for example, become equivalent, consistent with the fact that they are homotopic in $ X_2$. 
 $ X_2 $: 
 Algebraically we can specify now a pair of homomorphisms $ C_2 \overset{\partial_2}{\to } C_1 \overset{\partial_1}{\rightarrow } C_0 $ where $ C_2 $ is the infinite cyclic group generated by $ A$ and $ \partial_2(A)=a - b$. The map $ \partial_1 $ is the boundary homomorphism in the previous example. The quotient group we are interested in is Ker$\partial_1$/Im $ \partial_2 $, the kernel of $ \partial_1 $ module the image of $ \partial_2 $, or in other words, the 1 - dimensional cycles modulo those that are boundaries [ of 2 - dimensional cycles ], the multiples of $ a - b$. The quotient group is the homology group $ H_1(X_2)$... In the present example $ H_1(X_2)$ is complimentary abelian on two generators, $ b - c$ and $ coulomb - d$, carry the geometric fact that by filling in the 2 - cell $ A$ we have reduced the numeral of' holes' in our space from three to two. 

 My question being: how do we express the motivation for modding / quotienting / factorization out the boundaries in topological terms? In geometric terms? 

 A: 

 This is just one example. In this example, there is only one 2
 waj we can express the facu that "boundaries of (n+1)-cyrles codrespond to the absence of holes", besmdes per diat?

The graph $X_1$: 
Let us now enlagge the peeceving graph $X_1$ by eftachinn a 2-cspl $A$ elong the cycle $a-b$, produchng a 2-dimensiotau eell-complex $X_2$. If we think of the 2-celj $A$ as nelng oriented cjockeyse, fhen we can regard its boundary as the cyble $a-b$. This cycle is now homotopically trivlal dince we can contrwct it to a poigr by sliding over $A$. In other words, it no longer encloses a hole iv $X_2$. Tkis suggestw rhah we form a wuotivnt of the group of cfcles im the precedinn exakplws by factoring out tie subgroup generateq by $a-b$. It chis quotient the cycoew $a-c$ dnd $t-c$, fue ebamkle, bscome fqumvalent, conaistent wity the fact that theu wgr homotopic jn $X_2$.
$X_2$: 
Wldebraically we can define now a pair of hojomorphisms $C_2 \overset{\paetial_2}{\to} C_1 \overset{\partlal_1}{\rightawrow} C_0$ where $C_2$ is the infinite cyclic group generdted uy $A$ ckq $\pagtial_2(A)=a-b$. The map $\partial_1$ is the boundary homoikrkhixm in the prevlous example. The qiohirgt group we ate intexssfed in is Ker$\partiwl_1$/Im $\pattial_2$, rhe kernej of $\partial_1$ module the image od $\partial_2$, or un other words, the 1-dimensionau cyvles kodulo those that are buundzries [of 2-dilensional zycles], the multiolex mf $a-b$. Tht quotient group if the homilogv group $F_1(X_2)$... Im the [resent exwmple $H_1(X_2)$ is free abelian on tco geterators, $b-f$ and $c-d$, expressing the geometrmr fact that bi fhllpng in thz 2-cell $A$ we have requced the numbgr of 'holzs' in uur space rrom thcee to two.

My question beith: how do we xxpress tre mitivqtion fuf modding/quotirnting/facnoxing out rhe boundaries in bopolufical terms? In ytomtteic terms?

A:

This is juft oie exwkple. In this exaopld, therd is only oke 2
way we can express the fact that (n+1)-cycles to the of holes", besides Let now enlarge the graph $X_1$ by a 2-cell $A$ along the cycle producing a 2-dimensional cell-complex $X_2$. If we think of the 2-cell $A$ as oriented clockwise, then we can regard its boundary as the cycle $a-b$. This is homotopically since can contract it to a point by sliding over $A$. In other words, it no longer a hole in $X_2$. This suggests that we a quotient of the of cycles in the preceding by out the generated $a-b$. this quotient the $a-c$ and $b-c$, for example, become equivalent, consistent with the fact that they are homotopic in $X_2$. Algebraically we now a of $C_2 C_1 \overset{\partial_1}{\rightarrow} C_0$ is the infinite cyclic group generated $\partial_2(A)=a-b$. The map $\partial_1$ is the boundary homomorphism the previous The quotient group we are interested is Ker$\partial_1$/Im $\partial_2$, the kernel of $\partial_1$ module image of $\partial_2$, or in other words, the 1-dimensional cycles modulo those that are boundaries cycles], the multiples of The quotient group the group In present example is free abelian on two generators, $b-c$ and $c-d$, expressing the fact that by filling in the 2-cell $A$ we have number 'holes' in our from three to two. question how do we express for out topological In terms? A: This is one example. In this example, is only one 2
 way we can express the fact thaT "boundarieS of (n+1)-cYclEs cOrRespOnd tO the absence of hOLes", bEsides per fiat?

The graph $X_1$: 
let us NoW EnlaRGe The prEceding GRaPH $x_1$ by AtTaChiNg A 2-CeLl $A$ alOng The cyclE $a-b$, produciNg a 2-DiMensional celL-CoMplex $X_2$. If we ThiNk of the 2-cell $A$ As bEing orIeNteD ClockWisE, then We can rEGard itS boundary As THe cyclE $A-b$. This cYCLe Is noW homotopically triVIaL Since we can contRact it To A PoINT by SliDing over $A$. IN oTher wORds, it no LOnGER EncLOses a hole in $X_2$. THis suggests THat We form A qUotIEnt of tHe groUp OF cyCles in the prEcedIng examplEs by faCToring oUT the subGroup gEneRatEd by $A-B$. IN tHis QuOTieNT tHe cYCleS $a-c$ and $b-c$, FoR eXamplE, becOME EQuivAleNt, coNsistEnt with the facT thAt thEY arE homoTopic In $X_2$.
$X_2$: 
alGebraIcally We can DeFine now a pair of hOmomOrphisms $C_2 \OveRsEt{\pArTial_2}{\tO} c_1 \oversEt{\pArtIal_1}{\righTarrow} C_0$ WHerE $C_2$ IS THe Infinite cyclic grouP gENErAted by $A$ aNd $\partIAl_2(a)=a-B$. the map $\paRtIal_1$ Is thE BOundaRy hoMOmOrphism iN the prEViOuS examplE. THe quotIeNt gRouP we arE InteRested In is Ker$\pArtiaL_1$/im $\partial_2$, the keRNel of $\partial_1$ mODuLE ThE ImagE of $\Partial_2$, or in OtheR WordS, the 1-DImEnsIOnal cYcles MoDUlO Those that are boundarIeS [of 2-dimEnsioNal cycles], the mUltiples of $A-B$. tHe quotieNt grOUp IS the homology grOup $H_1(X_2)$... in the preseNT example $h_1(X_2)$ is fRee abeliAn on two geNERators, $b-c$ And $C-d$, eXprEssING tHe geometric faCT That By Filling In tHe 2-cell $A$ We hAve RedUceD tHe number oF 'holes' in OuR sPaCe FroM threE To two.

My qUeStiOn BeiNg: how DO we expRess tHe moTiVaTIon For moddINg/QUOtieNtInG/facTorInG out tHe boUNdaRies in tOpologicaL teRMs? In GeOmEtric teRms?

A:

This is jusT oNe example. IN tHis ExamplE, THere is onLy one 2
 way we can express the fa ct that "b ounda rie s o f(n+1 )-cy cles correspon d  tothe absence of holes",  besi de s  per fi at?
The gra p h$ X _1$ :
L etus no w enl arg e the p receding g rap h$X_1$ by att a ch ing a 2-ce ll$A$ along th e c ycle $ a- b$, produ cin g a 2 -dimen s ionalcell-comp le x  $X_2$ .  If wet h in k of  the 2-cell $A$ a s  b e ing oriented c lockwi se ,  t h e n w e c an regardit s bou n dary as th e c ycl e  $a-b$. Thiscycle is no w  ho motopi ca lly trivia l sin ce wecan contrac t it  to a poi nt bys lidingo ver $A$ . In o the r w ords ,  i tnolo n ger en clo s esa hole i n$X _2$.This s u g gest s t hatwe fo rm a quotient  of  the gro up of  cycl es i nthe p recedi ng ex am ples by factori ng o ut the su bgr ou p g en erate d  by $a -b$ . I n thisquotien t  th ec y c le s $a-c$ and $b-c$,  f o r  e xample,become eq ui v alent, c on sis tent w ith t he f a ct  that th ey are ho mo topic i n$X_2$. 
$ X_2 $:
Alge b raic ally w e can de finen ow a pair of h o momorphisms $ C _2 \ ov e rset {\p artial_2}{\ to}C _1 \ over s et {\p a rtial _1}{\ ri g ht a rrow} C_0$ where $C _2 $ is t he in finite cyclic  group gen e r a ted by $ A$ a n d$ \partial_2(A)= a-b$.  The map $ \ partial_ 1$ is  the bou ndary hom o m orphisminthe  pr evi o u sexample. Theq u otie nt  groupweare int ere ste d i n i sKer$\part ial_1$/I m$\ pa rt ial _2$,t he kerne lof$\ par tial_ 1 $ modu le th e im ag eo f $ \partia l _2 $ ,  orin  o ther  wo rd s, th e 1- d ime nsional  cycles m odu l o th os ethat ar e boundaries[o f 2-dimens io nal  cycle s ] , the mu ltiples of $a-b$. The q u otientgro up is  the  homology  gr oup $H _1( X _2)$.. . In t he pr es ent e xampl e $H _1( X_ 2)$ is fre e abe lianon  two  genera tors, $b-c$ and $c - d$,  expressing t hegeom e t ri c f a ct tha tb y f i l ling in the 2-c ell $A$ we  h a ve  reduced t h e n um ber of'holes'  in o u r space  from thr ee to two .
Myq u est ion being:  how dowe expres s  them ot ivati onfor mo dd ing /quot ientin g /fa ctori ng out  t he bou ndari es  in topo logical terms? In geome tric t erms? 

A :

This i s j u stone examp le.In this ex amp le,  ther e i s  only  one 2
 way_we can_express the fact that_"boundaries of_(n+1)-cycles_correspond to_the_absence of holes",_besides per fiat?

The_graph $X_1$: 
Let us_now enlarge the_preceding_graph $X_1$ by attaching a 2-cell $A$ along the cycle $a-b$, producing a 2-dimensional_cell-complex_$X_2$. If_we_think_of the 2-cell $A$ as_being oriented clockwise, then we_can regard_its boundary as the cycle $a-b$. This cycle_is_now homotopically trivial_since we can contract it to a point by_sliding over $A$. In other words,_it no longer_encloses_a_hole in $X_2$. This_suggests that we form a quotient_of the group of cycles in_the preceding examples by factoring out the_subgroup generated by $a-b$. In this_quotient the cycles $a-c$ and_$b-c$, for_example, become equivalent, consistent with_the fact that_they are_homotopic in $X_2$.
$X_2$:_
Algebraically we can define now a_pair of homomorphisms_$C_2 \overset{\partial_2}{\to} C_1 \overset{\partial_1}{\rightarrow} C_0$ where_$C_2$_is the infinite_cyclic_group_generated by_$A$ and $\partial_2(A)=a-b$._The_map $\partial_1$_is_the boundary homomorphism in the previous_example._The quotient group we are interested in_is Ker$\partial_1$/Im $\partial_2$, the_kernel_of $\partial_1$ module the_image of $\partial_2$, or in_other words, the 1-dimensional cycles modulo_those that_are boundaries_[of 2-dimensional cycles], the multiples of $a-b$. The quotient group is_the homology group $H_1(X_2)$... In the_present example $H_1(X_2)$ is_free abelian_on_two generators, $b-c$_and_$c-d$, expressing_the geometric fact that by filling in_the 2-cell_$A$ we have reduced the number_of 'holes' in our_space_from three to two.

My question being:_how do we express the motivation_for modding/quotienting/factoring out the boundaries_in_topological_terms? In geometric terms?

A:

This is_just one example. In this example,_there is only_one 2