In an exercise on Flasque sheaves, I used Snake lemma. So, I thought it is better to mention it separately with proof. Lemma : Given a commutative diagram as below we have exact sequence Proof : This is just question of diagram chasing. It is good if one can prove this on their own with out looking for proof from some other source. Let \(a\in \text{Ker}(f)\) i.e., \(f(a)=0\) which then imply \(v_1(f(a))=0\) which is same as saying \(g(u_1(a))=0\) (as the first square is commutative) i.e., \(u_1(a)\in \text{Ker}(g)\). So, we have map \(\tilde{u}_1:\text{Ker}(f)\rightarrow \text{Ker}(g)\) given by \(a\mapsto u_1(a)\). For similar reasons, \(b\mapsto u_2(b)\) gives map \(\tilde{u}_2:\text{Ker}(g)\rightarrow \text{Ker}(h)\). As \(\tilde{u}_1\) and \(\tilde{u}_2\) are just restrictions of \(u_1\) and \(u_2\), it follows that the sequence \(0\rightarrow \text{Ker(f)}\xrightarrow{\tilde{u}_1} \text{Ker(g)}\xrightarrow{\tilde{u}_2} \text{Ker(h)}\) is an exact sequence. Define \(\tilde{v}_1:\text{Coker(f)}\rightarrow \text{Coker(g)}\) by \(a+f(M_1)\rightarrow v_1(a)+g(M_2)\). Let \(a_1+f(M_1)=a_2+f(M_2)\), then, \(a_1-a_2\in f(M_1)\) i.e., \(a_1-a_2=f(m)\) for some \(m\in M_1\). So, we have \(v_1(a_1-a_2)=v_1(f(m))=g(u_1(m))\in g(M_2)\), thus, \(v_1(a_1)-v_1(a_2)\in g(M_2)\) i.e., \(v_1(a_1)+g(M_2)=v_1(a_2)+g(M_2)\) i.e., \(\tilde{v}_1(a_1+f(M)_1)=\tilde{v}_1(a_2+f(M_2))\). Thus, \(\tilde{v}_1:\text{Coker(f)}\rightarrow\text{Coker(g)}\) given by \(a+f(M_1)\rightarrow v_1(a)+g(M_2)\) is well defined. Similarly, \(\tilde{v}_2:\text{Coker(g)}\rightarrow \text{Coker(h)}\) given by \(b+g(M_2)\rightarrow v_2(b)+h(M_3)\) is a well defined map. As these are coming from \(v_1,v_2\) the sequence \(\text{CoKer}(f) \xrightarrow{\tilde{v}_1} \text{CoKer}(g) \xrightarrow{\tilde{v}_2} \text{CoKer(h)}\rightarrow 0\) is an exact sequence. Now, we define (connecting map) \(d:\text{Ker(h)}\rightarrow \text{Coker(f)}\) and show that this map connects the two exat sequences \(0\rightarrow \text{Ker(f)}\xrightarrow{\tilde{u}_1} \text{Ker(g)}\xrightarrow{\tilde{u}_2} \text{Ker(h)}\) and \(\text{CoKer}(f) \xrightarrow{\tilde{v}_1} \text{CoKer}(g) \xrightarrow{\tilde{v}_2} \text{CoKer(h)}\rightarrow 0\) giving the required exact sequence Let \(a\in \text{Ker(h)}\) i.e., \(h(a)=0\). As \(u_2\) is surjective, \(a=u_2(m_2)\) for some \(m_2\in M_2\). So, \(0=h(a)=h(u_2(m_2))=v_2(g(m_2))\). So, \(g(m_2)\in \text{Ker}(v_2)=\text{Im}(v_1)\). So, \(g(m_2)=v_1(n_1)\) for some \(n_1\in N_1\). Define \(d:\text{Ker(h)}\rightarrow \text{Coker(f)}\) as \(a\mapsto n_1+f(M_1)\) chosen as above. We prove that this is well defined. Let \(a=u_2(m_2)=u_2(m_2')\). Then, \(m_2-m_2'\in \text{Ker}(u_2)=\text{Im}(u_1)\). So, \(m_2-m_2'=u_1(m_1)\). Then, \(g(m_2)-g(m_2')=g(u_1(m_1))=v_1(f(m_1))\) i.e., \(v_1(n_1)-v_1(n_1')=v_1(f(m_1))\). As \(v_1\) is injective, this means \(n_1-n_1'=f(m_1)\in f(M_1)\) i.e., \(n_1+f(M_1)=n_1'+f(M_1)\). Thus, \(d:\text{Ker(h)}\rightarrow \text{Coker(f)}\) defined as \(a\mapsto n_1+f(M_1)\) is well defined. We have used surjectivity of \(u_2\) to define the map \(d\) and used injectivity of \(v_1\) to prove that it is well defined. I will write proof some other time that the resulting map is exact at \(\text{Ker}(h)\) and \(\text{Coker}(f)\).    

Definition : Let \(X\) be a topological space, \(\mathcal{F},\mathcal{G}\) be presheaves on \(X\). A morphism \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) is a collection \(\{\varphi(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)\}\) indexing over all open \(U\subseteq X\) such that the following diagram is commutative for open \(U\subseteq V\subseteq X\). Morphism of sheaves inducing Morphism of stalks : We see that any morphism of sheaves \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) induces morphism of stalks \(\varphi_p:\mathcal{F}_p\rightarrow \mathcal{G}_p\) for each \(p\in X\). Fix \(p\in X\). Let us define \(\varphi_p:\mathcal{F}_p\rightarrow \mathcal{G}_p\) i.e., for \((U,s)\in \mathcal{F}_p\) we give an open set \(V\) containing \(p\) and a section \(t\in \mathcal{G}(V)\) giving an element \((V,t)\in \mathcal{G}_p\). One obvious choice of an open set containing \(p\) is \(U\). For this \(U\), we have \(\varphi(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)\) sending \(s\) to \(\varphi(U)(s)\in \mathcal{G}(U)\). Define \(\varphi_p:\mathcal{F}_p\rightarrow \mathcal{G}_p\) as \((U,s)\mapsto (U,\varphi(U)(s))\). Map is well defined : We prove \((U,s) \sim (V,t)\) implies \((U,\varphi(U)(s))\sim (V,\varphi(V)(t))\). As \((U,s) \sim (V,t)\) there exists an open subset \(W\subseteq U\cap V\) containing \(p\) such that \(s|_W=t|_W\). We prove that \(\varphi(U)(s)|_W=\varphi(V)(t)|_W\) which implies \((U,\varphi(U)(s))\sim (V,\varphi(V)(t))\). The commutative diagram gives \(\varphi(U)(s)|_W=\varphi(W)(s|_W)\). Similar diagram in which \(U\) is replaced by \(V\) gives that \(\varphi(V)(t)|_W=\varphi(W)(t|_W)\). As \(s|_W=t|_W\) we have \(\varphi(W)(s|_W)=\varphi(W)(t|_W)\), concluding that \(\varphi(U)(s)|_W=\varphi(V)(t)|_W\). So, given a morphism of sheaves \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) we have well defined morphism \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) for each \(p\in X\). Theorem : A morphism of sheaves \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) is an isomorphism of sheaves iff the induced map \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) is an isomorphism for each \(p\in X\). Proof : Let \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) is an isomorphism of sheaves i.e., \(\varphi(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)\) is an isomorphism of groups for each open \(U\subseteq X\). Fixing \(p\in X\) we prove that \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) is an isomorphism. Let \((U,s)\in \mathcal{F}_p\) be such hat \((U,\varphi(U)(s))=0\in \mathcal{G}_p\) i.e., \(\varphi(U)(s)|_W=0\) for some open \(W\subseteq U\). We thus have \(\varphi(W)(s|_W)=\varphi(U)(s)|_W=0\). As \(\varphi(W):\mathcal{F}(W)\rightarrow \mathcal{G}(W)\) is injective, this means \(s|_W=0\). Thus, \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) is an injective map. Let \((V,t)\in \mathcal{G}_p\) i.e., \(p\in V\) and \(t\in \mathcal{F}(V)\). As \(\varphi(V):\mathcal{F}(V)\rightarrow \mathcal{G}(V)\) is surjective, there exists \(s\in \mathcal{F}(V)\) such that \(\varphi(V)(s)=t\). So, \(\varphi_p((V,s))=(V,\varphi(V)(s))=(V,t)\). Thus, \(\varphi_p\) is surjective. So, \(\varphi:\mathcal{F}\rightarrow \mathcal{G}\) is an isomorphism of sheaves implies \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) is an isomorphism for each \(p\in X\). Conversely, suppose that \(\varphi_p:\mathcal{F}_p\rightarrow\mathcal{G}_p\) is an isomorphism for each \(p\in X\). We prove \(\varphi(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)\) is an isomorphism for each open \(U\subseteq X\). Fix \(U\subseteq X\) and consider \(\varphi(U):\mathcal{F}(U)\rightarrow\mathcal{G}(U)\). Let \(s\in \mathcal{F}(U)\) be such that \(\varphi(U)(s)=0\). Fix \(p\in U\) and consider \(\varphi_p:\mathcal{F}_p\rightarrow \mathcal{G}_p\). We have \((U,s)\in \mathcal{F}_p\) with \(\varphi_p((U,s))=(U,\varphi(U)(s))=0\). As \(\varphi_p\) is injective, this means that \(s|_{W_p}=0\) for some \(W_p\subseteq U\) containing \(p\). This is true for all \(p\in U\). So, we have an open cover \(\{W_p\}_{p\in U}\) of \(U\) and \(s\in \mathcal{F}(U)\) such that \(s|_{W_p}=0\). Identity axiom of sheaf implies that \(s=0\). So, \(\varphi(U):\mathcal{F}(U)\rightarrow \mathcal{G}(U)\) is injective. Let \(s\in \mathcal{G}(U)\). Fix \(p\in U\) and consider \(\varphi_p:\mathcal{F}_p\rightarrow \mathcal{G}_p\). As \((U,s)\in \mathcal{G}_p\) and \(\varphi_p\) is surjective, there exists \((V,t_p)\in \mathcal{F}_p\) such that \(\varphi_p((V,t_p))=(U,s)\) i.e., \((V,\varphi(V)(t_p))=(U,s)\in \mathcal{F}_p\) i.e., \(s|_{W_p}=\varphi(V)(t_p)|_{W_p}\) for some \(p\in W_p\subseteq U\cap V\). Idea is to glue the sections \(t_p|_{W_p}\in \mathcal{F}(W_p)\) to get a section \(t\in \mathcal{F}(U)\). For that we show that \(t_p|_{W_p\cap W_q}=t_q|_{W_p\cap W_q}\). We have the following commuative diagram,which says that \(\varphi(W_p)(t_p|_{W_p})|_{W_p\cap W_q}=\varphi(W_p\cap W_q)(t_p|_{W_p\cap W_q}).\) ...

Let \(A\) be a ring. We have corresponding topological space \(X=\text{Spec}(A)\), the collection of all prime ideals of \(A\) with Zariski Topology. We now define a sheaf on \(X\) called the structure sheaf, denoted by \(\mathcal{O}_X\). This \(X\) with this structure sheaf \(\mathcal{O}_X\) is called an affine scheme, These affine schemes are building blocks of what is called an arbitrary scheme. To define a sheaf on \(X\) we need to associate a ring for each \(U\) open in \(X\). We do that as follows :where the condition \(\dagger\) says that given \(p\in U\) we have \(s(p)\in A_p\) and that \(s\) is locally a fraction i.e., given \(p\in U\) there exists an open set \(U(p)\subseteq U\) and \(a\in A, f\in A\) such that \(s(q)=\frac{a}{f}\in A_q\) for all \(q\in U(p)\). The verification that this gives a sheaf on \(X\) is same as that of the verification that sheafification of a sheaf is a sheaf. We can see the similarity between the definitions. More details can be found here about the similarity. So, \((X,\mathcal{O}_X)\) forms a ringed space, which we call an affine scheme. We will now see results about the global sections, stalks of structure sheaf and what does structure sheaf give on basic open subsets of \(X=\text{Spec}(A)\). Proposition : Let \(A\) be a ring, and \((\text{Spec A},\mathcal{O})\) its spectrum. For any \(\mathfrak{p}\in \text{Spec A}\), the stalk \(\mathcal{O}_{\mathfrak{p}}\) of the sheaf \(\mathcal{O}_{}\) is isomorphic to the local ring \(A_{\mathfrak{p}}\) i.e., \(\mathcal{O}_{\mathfrak{p}}\cong A_{\mathfrak{p}}\). For any element \(f\in A\), the ring \(\mathcal{O}(D(f))\) is isomorphic to the localized ring \(A_f\) i.e., \(\mathcal{O}(D(f))\cong A_f\). In particular, \(\Gamma(\text{Spec A}, \mathcal{O})\cong A\). Proof : Let \(\mathfrak{p}\in X\). We define a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) and show that this is a bijection. Defining the map - Let \([(U,s)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., \(U\) is an open set in \(X\) containing \(p\) and \(s\in \mathcal{O}(U)\). By definition, \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\). To get an element in \(A_{\mathfrak{p}}\) given \(s\), its only natural to consider image of \(\mathfrak{p}\) under \(s\) namely \(s(\mathfrak{p})\in A\). Defining \(s\mapsto s(\mathfrak{p})\) gives a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\). Showing that the map is well defined - Suppose \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., there is an open set \(W\subset U\cap V\) containing \(\mathfrak{p}\) such that \(s|_{W}=t|_W\). As \(\mathfrak{p}\in W\), we have in particular \(s(\mathfrak{p})=t(\mathfrak{p})\). So, there is a well defined map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\). Showing that the map is Injective - For \([(U,s)],[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) with \(s(\mathfrak{p})=t(\mathfrak{p})\), we show that \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\). As \(\mathfrak{p}\in U\), for \(s: U\rightarrow\bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) there exists open \(U(\mathfrak{p})\subset U\) containing \(\mathfrak{p}\) and \(a,f\in A\) such that \(s(\mathfrak{q})=\frac{a}{f}\) for all \(\mathfrak{q}\in U(\mathfrak{p})\). Similarly, for \(t:V\rightarrow\bigsqcup_{\mathfrak{q}\in V}A_{\mathfrak{q}}\) there exists open \(V(\mathfrak{q})\subseteq V\) and \(b,g\in A\) such that \(t(\mathfrak{q})=\frac{b}{g}\) for all \(\mathfrak{q}\in V(\mathfrak{p})\). In particular, \(\frac{a} {f}=s(\mathfrak{p})=t(\mathfrak{p})=\frac{b}{g}\). Let \(\mathfrak{q}\in U(\mathfrak{p})\cap V(\mathfrak{p})\). Then, \(s(q)=\frac{a}{f}=\frac{b}{g}=t(q)\). Thus, we have \(s|_{U(\mathfrak{p})\cap V(\mathfrak{q})}=t|_{U(\mathfrak{p})\cap V(\mathfrak{q})}\). Thus, \([(U,s)]=[(V,t)]\). So, \(s\mapsto s(\mathfrak{p})\) is injective. Showing that the map is surjective - Let \(\frac{a}{f}\in A_{\mathfrak{p}}\), we want to choose an open set \(U\) containing \(\mathfrak{p}\) and \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) such that \(s(\mathfrak{p})=\frac{a}{f}\). One choice for \(s\) is sending \(q\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). For this, we need \(f\notin\mathfrak{q}\) i.e., \(\mathfrak{q}\in D(f)\). Let \(U=D(f)\) and consider \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) sending \(\mathfrak{q}\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). We then have \(s(\mathfrak{p})=\frac{a}{f}\in A_{\mathfrak{p}}\). Thus, the map is surjective. So, we have isomorphism \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) given by \(s\mapsto s(\mathfrak{p})\). Let \(f\in A\). We define a map \(A_f\rightarrow \mathcal{O}(D(f))\) and show that this is a bijection. Defining the map - Given \(\frac{a}{f}\in A_f\) we assign \(s\in \mathcal{O}(D(f))\) where \(s:D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\). Let \(q\in D(f)\) then, \(f\notin q\). So, \(\frac{a}{f}\) is defined in \(A_q\). So, define \(s(q)\) to be the image of \(\frac{a}{f}\) in \(A_q\) for each \(q\in D(f)\). It is clearly a well defined function. Similarly we define for \(\frac{a}{f^n}\in A_f\) a map \(s:D(f^n)=D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\) as \(q\mapsto \frac{a}{f^n}\in A_q\). Showing that the map is injective - Suppose \(\frac{a}{f^n},\frac{b}{f^m}\in A_f\) is such that the corresponding maps \(s,t\) are equal i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_q~\forall q\in D(f)\) i.e., given \(q\in D(f)\) there exists \(t_q\notin q\) such that \(t_q(af^m-bf^n)=0\). Consider the case when \(D(f)=\{q\}\). As \(t\notin q\), we have \(q\in D(t)\) i.e., \(D(f)\subseteq D(t)\) i.e., \(V(t)\subseteq V(f)\) i.e., \(\sqrt{(f)}\subseteq \sqrt{(t)}\). As \(f\in \sqrt{(f)}\) we have \(f^l=td\) for some \(d\in A\). We have \(t(af^m-bf^n)=0\) which implies \(td(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\) i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\) and we are done. Suppose \(D(f)=\{q_i\}_{i\in \Lambda}\). As \(t_i\notin q_i\) we have \(q_i\in D(t_i)\) i.e., \(D(f)\subseteq \bigcup_{i\in \Lambda} D(t_i)\). As in previous observation, this means \(f^l\) is in the ideal generated by \(\{t_i\}\) for some \(l\in \mathbb{N}\). So, we have (after rearranging indices in \(\Lambda\)) \(f^l=a_1t_1+\cdots+a_nt_n\) for some \(a_i\in A\). As \(t_i(af^m-bf^n)=0\), we have \(a_it_i(af^m-bf^n)=0\) for all \(i\). So, \(\sum_{i=1}^na_it_i(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\). Thus, \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\). Thus, the map \(A_f\rightarrow \mathcal{O}(D(f))\) is injective.  

This is an exercise from Hartshorne's Algebraic Geometry book. A part of this exercise is called QcQs lemma in Ravi Vakil's Foundations of Algebraic Geometry notes.