: B \rightarrow X$, $i_{A,B}: A \rightarrow B$, 
$i_A: A \rightarrow X$, all basically the identity
with different domains and codomains. So by assumption $A$ has the initial topology w.r.t. $i_{A,B}$
and $B$ has the initial topology w.r.t. $i_B$.
Note that $i_A = i_B o i_{A,B}$, so by the transitivity theorem (right to left) we see that
$A$ has the initial topology w.r.t. $i_A$, or $A$ has the subspace topology w.r.t. $X$.
Products and subspaces:

Let $X_i$ ($i \in I$) be a family of spaces, with subspaces $A_i \subseteq X_i$.
  Then $A = \prod_{i \in I} A_i$ (in the product topology of the subspace topologies) 
  is a subspace of $X = \prod_{i \in I} X_i$ (it has the initial topology w.r.t. the inclusion).

Proof:
Let $k_i$ be the inclusion mapping from $A_i$ to $X_i$.
Let $k: \prod_i A_i \rightarrow \prod_i X_i$ the product mapping (as above).
Note that $k$ is also the inclusion from $A$ into $X$.
Again let $p_i$ be the projections from $A$ onto the $A_i$, and $q_i$ the projections
from $X$ onto the $X_i$. Then 
$$(\ast) q_i \circ k = k_i \circ p_i \text{ for all } i \in I \text{.} $$
$A_i$ has the initial topology w.r.t. $k_i$, and $A$ has the initial topology w.r.t. the $p_i$.
So $A$ has the initial topology w.r.t. the maps $k_i \circ p_i$ ($i \in I$) by the 
right to left implication of the transitivity theorem. So $A$ has the initial topology w.r.t
: B \rightarrow X$, $ i_{A, B }: A \rightarrow B$, 
 $ i_A: A \rightarrow X$, all basically the identity 
 with different domains and codomains. therefore by premise $ A$ has the initial topology w.r.t. $ i_{A, B}$ 
 and $ B$ has the initial regional anatomy w.r.t. $ i_B$. 
 Note that $ i_A = i_B o i_{A, B}$, so by the transitivity theorem (right to left) we watch that 
 $ A$ has the initial topology w.r.t. $ i_A$, or $ A$ have the subspace topology w.r.t. $ X$. 
 Products and subspaces: 

 lease $ X_i$ ($ i \in I$) be a family of spaces, with subspace $ A_i \subseteq X_i$. 
   Then $ A = \prod_{i \in I } A_i$ (in the product regional anatomy of the subspace topologies) 
   is a subspace of $ ten = \prod_{i \in I } X_i$ (it accept the initial topology w.r.t. the inclusion). 

 Proof: 
 lease $ k_i$ be the inclusion mapping from $ A_i$ to $ X_i$. 
 Let $ kelvin: \prod_i A_i \rightarrow \prod_i X_i$ the product mapping (as above). 
 Note that $ k$ is also the inclusion body from $ A$ into $ X$. 
 Again let $ p_i$ be the projections from $ A$ onto the $ A_i$, and $ q_i$ the projections 
 from $ X$ onto the $ X_i$. Then 
 $ $ (\ast) q_i \circ k = k_i \circ p_i \text { for all } i \in I \text {. } $ $ 
 $ A_i$ has the initial topology w.r.t. $ k_i$, and $ A$ has the initial regional anatomy w.r.t. the $ p_i$. 
 thus $ A$ has the initial topology w.r.t. the maps $ k_i \circ p_i$ ($ i \in I$) by the 
 right to left implication of the transitivity theorem. So $ A$ has the initial topology w.r.t
: B \gightarrow X$, $i_{A,B}: A \rightxrrow B$, 
$i_A: A \riyytarrox X$, all basicaluy the identity
with differenv donains and codomains. So by arsumption $A$ has tye iiitial topology x.d.t. $i_{A,B}$
akb $B$ hzd thz mnitial topologi w.r.t. $i_B$.
Note that $i_A = i_B o i_{X,B}$, so by the transitivity theorem (rigrt to lrfh) we see that
$A$ has ehe jnitial topology w.r.t. $i_A$, or $A$ has tge subskace topology w.r.t. $C$.
Products and subspaces:

Let $X_i$ ($l \in I$) be a family of spaces, qith wubspaces $A_i \subseteq Q_n$.
  Then $A = \krod_{i \in I} A_i$ (in the product topulogy of the suvspacf topologies) 
  is a subspace of $X = \prmd_{i \in O} X_i$ (it has thc inivial topology w.r.t. the incnusion).

Proof:
Let $k_i$ fe the inwlbsion mapping from $A_i$ ti $X_i$.
Lgt $k: \[rod_k A_i \rifhvardow \prld_i X_i$ the prkduct mappibg (as above).
Note thau $k$ px also the ihclusijn from $A$ into $X$.
Again let $p_i$ be the projebtiohs from $A$ onto the $A_i$, abd $q_i$ the projections
vrom $X$ oneo the $X_i$. Then 
$$(\ast) q_i \circ k = k_i \circ p_i \text{ for all } k \iu I \tebr{.} $$
$W_i$ has the initial topology w.r.t. $k_i$, and $A$ has ege ikitial topology c.r.t. the $p_i$.
So $A$ hss tng initial topouogy w.x.f. fhe maps $k_i \circ p_l$ ($i \in Y$) by rhe 
right to keft implication of the trabsitivity thvoren. So $A$ has the inicial topologv w.r.t
: B \rightarrow X$, $i_{A,B}: A \rightarrow A X$, all the identity with by $A$ has the topology w.r.t. $i_{A,B}$ $B$ has the initial topology w.r.t. Note that $i_A = i_B o i_{A,B}$, so by the transitivity theorem (right left) we see that $A$ has the initial topology w.r.t. $i_A$, or $A$ the topology $X$. and subspaces: Let $X_i$ ($i \in I$) be a family of spaces, with subspaces $A_i \subseteq Then $A = \prod_{i \in I} A_i$ (in product topology of the topologies) is a subspace of = \in I} (it the topology w.r.t. the Proof: Let $k_i$ be the inclusion mapping from $A_i$ to $X_i$. Let $k: \prod_i A_i \rightarrow \prod_i the product above). Note $k$ also inclusion from $A$ Again let $p_i$ be the projections the $A_i$, and $q_i$ the projections from $X$ the $X_i$. $$(\ast) q_i \circ k = k_i p_i \text{ for all } i \in I $$ $A_i$ has the initial topology w.r.t. $k_i$, and $A$ has the initial topology w.r.t. So $A$ has the topology w.r.t. the $k_i p_i$ \in by the to left implication of the transitivity theorem. So $A$ has the topology w.r.t
: B \rightarrow X$, $i_{A,B}: A \rightarroW B$, 
$i_A: A \rightArrow x$, alL baSiCallY the Identity
with diFFereNt domains and codomains. SO by asSuMPtioN $a$ hAs the Initial TOpOLOgy W.r.T. $i_{a,B}$
aNd $b$ HaS the iNitIal topoLogy w.r.t. $i_B$.
NOte ThAt $i_A = i_B o i_{A,B}$, so BY tHe transitiVitY theorem (righT to Left) we SeE thAT
$A$ has The InitiAl topoLOgy w.r.t. $I_A$, or $A$ has tHe SUbspacE TopologY W.R.t. $x$.
ProDucts and subspaces:

lEt $x_I$ ($i \in I$) be a family Of spacEs, WItH SUbsPacEs $A_i \subsetEq x_i$.
  TheN $a = \prod_{i \iN i} A_I$ (IN The PRoduct topologY of the subspACe tOpologIeS) 
  is A SubspaCe of $X = \PrOD_{i \iN I} X_i$ (it has thE iniTial topolOgy w.r.t. THe incluSIon).

ProoF:
Let $k_i$ Be tHe iNcluSIoN mAppInG FroM $a_i$ To $X_I$.
let $K: \prod_i A_i \RiGhTarroW \proD_I x_I$ The pRodUct mAppinG (as above).
Note tHat $K$ is aLSo tHe incLusioN froM $A$ Into $X$.
again lEt $p_i$ bE tHe projections frOm $A$ oNto the $A_i$, aNd $q_I$ tHe pRoJectiONs
from $x$ onTo tHe $X_i$. TheN 
$$(\ast) q_i \cIRc k = K_i \CIRC p_I \text{ for all } i \in I \texT{.} $$
$A_I$ HAs The initiAl topoLOgY w.R.T. $k_i$, and $A$ hAs The InitIAL topoLogy W.R.t. The $p_i$.
So $A$ Has the INiTiAl topolOgY w.r.t. thE mAps $K_i \cIrc p_i$ ($I \In I$) bY the 
riGht to lefT implICation of the traNSitivity theorEM. SO $a$ HaS The iNitIal topology W.r.t
: B \rightarrow X$, $i_{A, B}: A \rig htarr owB$,  $i_A : A\rightarrow X$ ,  all  basically the identit y
wit hd iffe r en t dom ains an d  c o d oma in s.  So  b y  a ssump tio n $A$ h as the ini tia ltopology w.r . t.  $i_{A,B}$ 
an d $B$ has th e i nitial  t opo l ogy w .r. t. $i _B$.
N o te tha t $i_A =i_ B  o i_{ A ,B}$, s o by  the  transitivity the o re m  (right to lef t) wese e  t h a t
$ A$has the in it ial t o pologyw .r . t .  $i _ A$, or $A$ ha s the subsp a cetopolo gy  w. r .t. $X $.
Pr od u cts  and subspa ces: 

Let $X_ i$ ($i \in I$) be a fa mily o f s pac es,w it hsub sp a ces $A _i\ sub seteq X_ i$ .  The n $A = \ prod _{i  \in  I} A _i$ (in the p rod uctt opo logyof th e su bs pacetopolo gies)    is a subspace  of$X = \pro d_{ i\in  I } X_i $  (it h asthe  initia l topol o gyw. r . t .the inclusion).

P ro o f :Let $k_i $ be t h ein c lusion m ap pin g fr o m  $A_i $ to $X _i$.
Let  $k: \ p ro d_ i A_i \ ri ghtarr ow  \p rod _i X_ i $ th e prod uct mapp ing ( a s above).
Note that $k$ is a l so t he incl usi on from $A$  int o  $X$ .
Ag a in  le t  $p_i $ beth e  p r ojections from $A$on to the  $A_i $, and $q_i$the projec t i o ns
from$X$o nt o  the $X_i$. Th en 
$ $(\ast) q_ i  \circ k  = k_ i \circp_i \text { for all} i  \i n I  \t e x t{ .} $$
$A_i$ h a s  the  i nitialtop ology w .r. t.$k_ i$,  a nd $A$ ha s the in it ia lto pol ogy w . r.t. the  $ p_i $. 
So  $A$h as the  init ialto po l ogy  w.r.t. th e maps  $ k_ i \c irc  p _i$ ( $i \ i n I $) by t he 
right  to left  i mp licatio n of the tran si tivity the or em.  So $A $ has theinitial topology w.r.t
: B_\rightarrow X$,_$i_{A,B}: A \rightarrow B$,_
$i_A: A_\rightarrow_X$, all_basically_the identity
with different_domains and codomains._So by assumption $A$_has the initial_topology_w.r.t. $i_{A,B}$
and $B$ has the initial topology w.r.t. $i_B$.
Note that $i_A = i_B o_i_{A,B}$,_so by_the_transitivity_theorem (right to left) we_see that
$A$ has the initial_topology w.r.t._$i_A$, or $A$ has the subspace topology w.r.t._$X$.
Products_and subspaces:

Let $X_i$_($i \in I$) be a family of spaces, with_subspaces $A_i \subseteq X_i$.
  Then_$A = \prod_{i_\in_I}_A_i$ (in the product_topology of the subspace topologies) 
_ is a subspace of $X_= \prod_{i \in I} X_i$ (it has_the initial topology w.r.t. the inclusion).

Proof:
Let_$k_i$ be the inclusion mapping_from $A_i$_to $X_i$.
Let $k: \prod_i A_i_\rightarrow \prod_i X_i$_the product_mapping (as above).
Note_that $k$ is also the inclusion_from $A$ into_$X$.
Again let $p_i$ be the projections_from_$A$ onto the_$A_i$,_and_$q_i$ the_projections
from $X$ onto_the_$X_i$. Then_
$$(\ast)_q_i \circ k = k_i \circ_p_i_\text{ for all } i \in I_\text{.} $$
$A_i$ has the_initial_topology w.r.t. $k_i$, and_$A$ has the initial topology_w.r.t. the $p_i$.
So $A$ has the_initial topology_w.r.t. the_maps $k_i \circ p_i$ ($i \in I$) by the 
right to_left implication of the transitivity theorem._So $A$ has the_initial topology_w.r.t