If $f,g:(X,A)\to (Y,B)$ are homotopic functions under a homotopy $H:X\times [0,1]\to Y$ such that $H(a,t)\in B$ for all $a\in A$ and $t\in [0,1]$, then $f_*=g_*:H_n(X,A)\to H_n(Y,B)$.   This question has already been posted here  and here  but I still have many doubts, I would like to know where we are using that $f,g:(X,A)\to (Y,B)$ are homotopic in .  I think it is enough to show that $f,g:X/A\to Y/B$ are homotopic, because in this case we already know that $f_*=g_*:H_n(X/A)\to H_n(Y/B)$ where $H_n(X/A)=H_n(X,A)$ and $H_n(Y/B)=H_n(Y,B)$. Now if $f,g:(X,A)\to (Y,B)$ are homotopic, how do I prove that $f,g:X/A\to Y/B$ are homotopic? Thank you.
124236
Working with a homework problem:  $X$ and $Y$ are topological spaces and $A\subset X$, $B\subset Y$.  I have $f,g:(X,A)\to (Y,B)$ as homotopic maps.  I need to show that the induced maps:  $\hat{f},\hat{g}$ are homotopic.  My instinct says that the homotopy I should use is $\hat{F}:I\times (X/A)\to (Y/B)$ should be given by $\hat{F}(t,[x]) := [F(t,x)]$.  I showed this map is well defined without any problem.  Edit:  More precisely, $\hat{f}$ maps the pair $(t,[x])$, where $x$ is any element in $[x]\in X/A$, to $[y]$, where $y$ is any element in $[F(t,x)]$.  But I'm having difficulty showing the map is continuous.  If I take $V$ to be an open subset of $Y/B$, how can I use the continuity of $F$ to show that $\hat{F}$ is continuous?
31628
Let $g:\mathbf{R\rightarrow R}$ be a twice differentiable function.  a) Prove that if $a_1 &lt; a_2$ and $g(a_1)=g(a_2)=0$, then there exists a point $m \in (a_1,a_2)$ such that $g'(m)=0$.  b) Prove that if $a_1 &lt; a_2 &lt; a_3$ and $g(a_1)=g(a_2)=g(a_3)=0$, then there exists a point $n \in (a_1,a_3)$ such that $g''(m)=0$.  I am just curious about how to prove this very rigorously. My intuition tells me that both parts are true, but I have a hard time putting it down in writing. Any general tips about how I can learn to do this better are also much appreciated!
50967
I'm working through a real analysis textbook, so I don't want the full answer. I'm only looking for a  on this problem.  The book starts the proof like this: let $f$ be a bijection of $\mathbb{N}_m$ onto $A$, and let $g$ be a bijection of $\mathbb{N}_n$ onto $B$. Define $h$ on $\mathbb{N}_{m+n}$ as $h(i):=f(i)$ for $i=1, ..., m$, and $h(i):=g(i)$ for $i=m+1,...,m+n$. Show that $h$ is a bijection from $\mathbb{N}_{m+n}$ onto $A \cup B$.  Ok, so with that, I started out proving that $h$ is an injection into $A \cup B$. The usual method for this is to prove that if $h(i)=h(j)$, where $i,j=1,...,m+n$, then $i=j$. I think I proved this by looking at four cases.   $1 \leq i,j \leq m$. In this case, $h(i)=f(i)$ and $h(j)=f(j)$, so because $f$ is a bijection, we know that $f(i)=f(j)$ implies that $i=j$. $ 1 \leq i \leq m$ and $m+1 \leq j \leq m+n$. We know that $i \neq j$ (obviously), so we want to prove that in this case, $h(i) \neq h(j)$. By definition, $h(i)=f(i)$ and $h(j)=g(j-m)$. Because $f$ is a bijection into $A$ and $g$ is a bijection into $B$, and $A \cap B = \emptyset$, we know that $f(i) \neq g(j-m)$, so $h(i) \neq h(j)$. This case is identical to case 2, except that $m+1 \leq i \leq m+n$ and $1 \leq j \leq m$, so the logic is reversed.  The final case, where $m+1 \leq i,j \leq m+n$, is similar to case 1, because since $h(i)=g(i)$ and $h(j)=g(j)$, we can use the fact that $g$ is a bijection to prove that $g(i)=g(j)$ implies that $i=j$.   This is where I got stuck. I know I need to prove that $h$ is surjective, but am I on the right track for proving that it's injective? I think proving that $h$ is a surjection shouldn't be too hard, but I don't know if I'm approaching this part of the proof correctly.
75204
Problem:  Let T be a nonempty set and A(T) the set of all permuations of T. Show that A(T) is a group under the operation of composition of functions.  Permutation of the set T is the bijective function $f: T \rightarrow T $  For A(T) to be a group it must be closed, associate have an identity for every element and an inverse.  Since f is  bijection the composition of two bijective functions is itself bijective so A(T) is closed. Every bijective function has an inverse, and composition of functions is associative. What about the identity? Are we to take $ A(T) = S_n $ here, which would be assuming that $T$={$1, ..., n$}? In which case $ I = \begin{bmatrix}1&amp;2&amp;...&amp;n\\1&amp;2&amp;...&amp;n\end{bmatrix} $  Not sure what I'm supposed to be showing since $T$ and A(T) are arbitrarly specified here.
59235
This question came up while showing the composition of a metric with a certain other  function gives another metric. Suppose I have some metric space $(X,d)$ and a  continuous, non-decreasing function $f$ on the nonnegative reals. Moreover, suppose  that $f(x)=0$ iff $x=0$, and $f$ also satisfies the triangle inequality in that $f (x+y)\leq f(x)+f(y)$.   Using these properties, it is not difficult to show that $f\circ d$ is yet another  metric on $X$, so $(X,f\circ d)$ is also a metric space.   I notice the (open) balls given by the metrics are of form  $$ B_d(x,r)=\{y\in X\mid d(x,y)\lt r\} $$ and  $$ B_{f\circ d}(x,r)=\{y\in X\mid (f\circ d)(x,y)\lt r\}, $$  so it seems that the topologies generated by the base of open balls in each case  would probably be the same. I would like to see how one would go about showing the topologies given by these two metrics are indeed the same. Thanks for any insight.
4519
Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.  I need advise or correction if something is incorrect with my proof. Thank you beforehand.  We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$  Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$  $\implies g(f(a))=g(f(b))$  $\implies g\circ f(a)=g\circ f(b)$  $\implies a=b$ (by injectivity of $g\circ f$)  $\implies f(a)=f(b)$  $\implies x=y$  Would appreciate any correction in proof writing also!
83382
Let $f: K^\bullet \to L^\bullet$ be homotopy equivalence. From theory of triangulated categories it follows that $C(f)^\bullet$ is contractible. But how can i produce explicit formula for homotopy of $Id_{C(f)^\bullet}$ and $0_{C(f)^\bullet}$?
17750
I know that the fundamental group of $X = \mathbb R^2 \setminus \{(0,0)\}$ is the same as the fundamental group of the circle $Y = S^1$, namely $\mathbb Z$.  However, $X$ and $Y$ are not homotopic, i.e. we can't find continuous maps $f:X\to Y, g : Y \to X$ such that $f \circ g$ is homotopic to $id_Y$ and $g \circ f$ is homotopic to $id_X$.  I would like to prove it, but I don't really know how to do it. If such $f$ and $g$ existed, then it would be something like $f : x \mapsto x/\|x\|$, and $g(Y)$ has to be compact... I don't know how to continue.  Any hint would be appreciated. I apologize if this has already been asked.
61085
Prove that if $f:X\to Y$ is a surjective function between sets, then there must exist a function $g:Y\rightarrow X$ such that $f\circ g=1_Y$.   I know that the identity function is onto, and if $f$ has right inverse, then $f$ must be onto; although, I haven't seen a proof of this. A good hint would be nice.
70550
Possible Duplicate:          I'm studying the Atiyah's commutative algebra book and I realized that in the beginning of the book, the author says as one of the conditions to a map be a homomorphism is $f(1)=1$, I would like to know if I can left this condition behind, i. e., if $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ we can have $f(1)=1$.  Thanks
85131
If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.  I'm wondering if you can help me with this one?
79978
Is this proof correct? To prove $F(A)\cap F(B)\subseteq F(A\cap B)  $ for all functions $F$.  Let any number $y\in F(A)\cap F(B)$. We want to show $y\in F(A\cap B).$ Therefore, $y\in F(A)$ and $y\in F(B)$, by definition of intersection. By definition of inverse, $y=F(x)$ for some $x\in A$ and $x\in B$ And so, $y=F(x)$ for some $x\in A\cap B$ And therefore, $y\in F(A\cap B)$  I have a gut feeling deep down that something is wrong. Can anyone help me pinpoint the mistake? I am not sure why am I having so much problems with functions. Am I not thinking in the right direction? Sources : 2nd Ed, P219 9.60 = 3rd Ed, P235 9.12, 9.29 - Mathematical Proofs, by Gary Chartrand, P214 Theorem 12.4 - Book of Proof, by Richard Hammack, P257-258 - How to Prove It, by D Velleman.
53398
I'm new at algebraic topology and I'm working on proving the following theorem  Theorem. Let $(X, x_0)$ and $(Y, y_0)$ two pointed topological spaces such that there exist two continuous pointed maps $f : (X, x_0) \to (Y, y_0)$ and $g : (Y, y_0) \to (X, x_0)$ such that $g \circ f$ and the identity $id_X$ are homotopic, and $f \circ g$ and the identity $id_Y$ are homotopic. So, the maps $$     \begin{array}{ccccc}         f_{\ast} &amp; : &amp; \pi_1(X, x_0) &amp; \longrightarrow &amp; \pi_1(Y, y_0) \\         &amp; &amp; [\gamma] &amp; \longmapsto &amp; [f \circ \gamma]     \end{array} $$ and $$     \begin{array}{ccccc}         g_{\ast} &amp; : &amp; \pi_1(Y, y_0) &amp; \longrightarrow &amp; \pi_1(X, x_0) \\         &amp; &amp; [\sigma] &amp; \longmapsto &amp; [g \circ \sigma]     \end{array} $$ are isomorphisms.  To this end, we need the following proposition (which I managed to demonstrate)  Proposition. Let $X$ and $Y$ two topological spaces and $x_0 \in X$. Let $f_1 : X \to Y$ and $f_2 : X \to Y$ be two continuous homotopic maps via the map $h : X \times [0, 1] \to Y$. Let $\gamma : I \to Y$; $t \mapsto h(x_0, t)$. Then, for all loop $\delta : [0, 1] \to X$ based at $x_0$, we have $$     [\overline{\gamma} \ast (f_1 \circ \delta) \ast \gamma] = [f_2 \circ \delta] \in \pi_1(Y, f_2(x_0)) $$ where $\ast$ is the composition of paths, and $\overline{\gamma}$ is the inverse path of $\gamma$.  So here is the demonstration (and where I Will mention the step where I'm stuck)  Proof of the theorem. Let $h : X \times [0, 1] \to X$ the homotopy such that $h(x, 0) = x$ and $h(x, 1) = g \circ f (x)$, for all $x \in X$. Let $\gamma : [0, 1] \to X$, $t \mapsto h(x_0, t)$.  From the proposition above, we have  $$     [\overline{\gamma} \ast \delta \ast \gamma] = [g \circ f \circ \delta] \in \pi_1(X, x_0) $$ for all loop $\delta : [0,1] \to X$ based at $x_0$.  Note that $\gamma$ is a loop based on $x_0$: $\gamma(0) = h(x_0, 0) = x_0$ and $\gamma(1) = h(x_0, 1) = g(f(x_0)) = g(y_0) = x_0$. So we can write $$     [\overline{\gamma}] \ast [\delta] \ast [\gamma] = (g \circ f)_{\ast}([\delta]) \in \pi_1(X, x_0) $$ This proves that $(g \circ f)_{\ast} = id_{\pi_1(X, x_0)}$ but I don't know why since composition of homotopy classes is not commutative.  (The end of the proof is pretty obvious by using the fact that $(g \circ f)_{\ast} = g_{\ast} \circ f_{\ast}$.)  Remark. I'm not actually searching for another proof of the isomorphism of $f_*$ and $g_*$, I'm just wondering why we deduced that $(g \circ f)_* = id_{\pi_1(X, x_0)}$ and any help or hint would be great.  Thanks in advance.  Best regards.  K. Y.
118216
So I came across this exercise: Show that if $Y$ is path-connected then the set of homotopy classes $[I,Y]$ of maps of $I=[0,1]$ into Y has a single point.   But I am confused here. If $Y=S^1$ then not all loops are homotopic! I mean, isn't true that $\pi_1(Y)\subset [I,Y]$? Or maybe I don't understand the set $[I,Y]$ well!
113070
When $f$ and $g$ are invertible functions we have  $$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$$  I think we have this in every Group $(A,*,u,')$  $(a*b)=c$   $(a*b)*b'=a=c*b'$  $c'*a=c'*(c*b')=b'$   $(c'*a)*a'=b'*a'$      $$c'=(a*b)'=b'*a'$$   and in general we have   $(a_1*a_2*...*a_n)'=a_n'*...*a_2'*a_1'$  1)What is the specific name of this theorem?   2)There is something wrong with my derivation?  EDIT Can we see this as a property of a function $f:A\rightarrow A$ on a structure $(A,\circ)$:  $f(x\circ y)=f(y)\circ f(x)$  ?
89065
suppose that $f$ is a mapping from $\mathbb R^n \to \mathbb R$.  a positively homogeneous of degree $n$ and suppose $f_1, f_2, .... f_N$ are continuous for $a \neq 0$. then   $$\sum_{i=1}^n a_if_i(a)= n*f(a)$$   now, they asked me to prove this. (Which is a total no no to me)  I write my proof as;  $f$ is a mapping from $\mathbb R^n \to \mathbb R$  A homogeneous function of degree $n$ if for all $a \in \mathbb R^n$ and all $L&gt;0$, $f(La)=L^n\,f(a)$. let $a \in \mathbb R^n$ and define function $g$ be a mapping from $[0,\infty) \to \mathbb R$. so I write $g(L) = f(La)-L^nf(a)$. then $L≥0$ let $L=0$ then, $g(L)=0$ and therefore $g'(x) =0$?  then i don't know what's goes next. can anyone help me about this?
68746
Prove that every continuous function $f$ can be written $f = g−h$, where   $g$ and $h$ are nonnegative and continuous.   If I define $g=\max\{f,0\}$ and $h=\max\{0,-f\}$, do I need to prove that $g$ and $h$ are still continuous and if so how do I do that?